Topic Of Nhocnhoc, Maths.vn

Tue Nov 10, 2009 10:53 am, by Potla

Hello all, this is my contribution to mathlinks.ro, which is the full log file of Topic of nhocnhoc or namely nguoivn. The problems have been translated to english and then edited by me. Hope this compilation will help us all.

This entry was quite long at first so I am attaching the (un-revised edition) of the file Smile

* I will try to correct the alignment of the document when I get enough time. Actually I had been preparing this for more than a month Very Happy

Topic Of Nhocnhoc.pdf

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Garden

Wed Nov 04, 2009 8:58 pm, by Potla

Sorry for getting off-mathematics (all my posts until now are related to mathematics) , but do you like this image? It is of my friend's garden. I really like this so posting....... Smile

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Chebyshev's Inequality

Wed Oct 14, 2009 11:14 pm, by Potla

A direct consequence of rearrangement, Chebyshev is a nice inequality that is able to deduce nicer results and can be extremely powerful sometimes too. This can be proved by rearrangement so this holds for all real numbers.
Statement
We already know it from our previous post; if $a_i$ and $b_i$ are similarly sorted sequences then
$\sum_{i = 1}^n a_ib_i\geq \frac1n\sum a_i \sum b_i$
And if these sequences are oppositely sorted then
$\sum_{i = 1}^n a_ib_i\leq \frac1n\sum a_i \sum b_i$
Since we already know it's proof; I will directly go to applications of this inequality in various problems. We have already proven Iran 1996 using Chebyshev's inequality. I have proved an inequality in Mathematical reflections this issue using simple Chebyshev; and I will post the inequality and my solution using Chebyshev after the deadline (1st November 2009) Smile

.We also have a little introduction to Some applications because of my previous post.So now I will have a short introduction to this inequality by deducing some known easy inequalities at first.
Smile

Nessbitt's inequality
We will prove for $a,b,c > 0$ we have $\sum \frac a{b + c}\geq \frac32$ in another way.
We have, the sequences $a,b,c$ and $\frac1{b + c}; \frac1{c + a}; \frac1{a + b}$ are similarly sorted so from Chebyshev;
$\sum \frac a {b + c}\geq \frac 13 (a + b + c)\left(\frac1{b + c} + \frac1{c + a} + \frac1{a + b}\right)$
$\ge \frac 13(a + b + c) \left(\frac9{2(a + b + c)} = \frac 32$ (From Cauchy Schwartz inequality)
2. (Problem by Darij Grinberg) For $a,b,c > 0$ prove that
$\sum\frac {a}{(b + c)^{2}}\geq{{a + b + c}\over 4}$

Since sequences $a,b,c$ and $\frac {1}{(b + c)^{2}},\frac {1}{(c + a)^{2}},\frac {1}{(a + b)^{2}}$ are similarly sorted hence from Chebyshev;
$\sum\frac {a}{(b + c)^{2}}\geq\frac {1}3\sum a\sum\frac {1}{(b + c)^{2}}$
$\geq\frac {1}3\sum a{{9}\over{4(ab + bc + ca)}}$ (from Iran 1996)
$\geq\frac {a + b + c}{4}\frac {3}{ab + bc + ca}\geq{{a + b + c}\over{4}}$
Since $ab + bc + ca\leq{{(a + b + c)^{2}}\over{3}}}\leq 3$
So we are done.

Applications
$\boxed{1}$ Let $a,b,c$ denote the lengths of the three sides opposite to the three angles $A,B,C$ of a triangle.
Show that:
$\frac {\pi}{3}\leq\frac {aA + bB + cC}{a + b + c}\leq\frac {\pi}{2}$
Solution
Let $a \geq b \geq c$ so $A \geq B \geq C$ . By Chebyshev inequality we have:

$\frac {aA + bB + cC}{a + b + c}\geq \frac {(\frac {a + b + c}{3})(A + B + C)}{a + b + c} = \frac {A + B + C}{3} = \frac {\pi}...$

For the right inequality, note that:
$2aA + 2bB + 2cC\leq\pi(a + b + c)$
$2aA + 2bB + 2cC\leq aA + bA + cA + aB + bB + cB + aC + bC + cC$
$A(b + c - a) + B(c + a - b) + C(a + b - c)\geq0$ which is true for $a,b,c$ sides of a triangle.

$\boxed{2}$ For $a,b,c\ge0$ Prove that $\frac {a^{3}}{bc} + \frac {b^{3}}{ac} + \frac {c^{3}}{ab}\ge a + b + c$
Sequences $a,b,c$ and $\frac1{bc}; \frac1{ac}; \frac1{ab}$ are similarly sorted.Hence from Chebyshev we have

$LHS \geq \frac {1}{3}(a + b + c) \sum_{cyc}{\frac {a^2}{bc} \geq \frac {1}{3}(a + b + c)\frac {(a + b + c)^2}{ab + bc + ca} \...$ $\Box$
$\boxed{3}$ Let $a,b,c > 0$ ; prove that :
$\frac {(b + c)(b^2 + c^2)}{a} + \frac {(c + a)(c^2 + a^2)}{b} + \frac {(a + b)(a^2 + b^2)}{c}\geq 4(a^2 + b^2 + c^2)$
Solution
Observe that sequences $\frac {b + c}{a}; \frac {c + a}{b}; \frac {a + b}{c}$ and $(b^2 + c^2); (c^2 + a^2); (a^2 + b^2)$ are similarly sorted (verify Smile

). Hence we can apply Chebyshev on these to obtain:
$\sum\frac {b + c}{a}\cdot (b^2 + c^2)\geq \frac 13 \sum \frac {b + c}{a}\sum (a^2 + b^2)$

$\geq \frac 13 \cdot 6 \cdot 2(a^2 + b^2 + c^2) = 4(a^2 + b^2 + c^2)}$
Because from AM-GM we have $\frac {b + c}{a} = \sum_{sym}\frac ab \geq 6$ .
Hence we are done Smile

$\Box$
$\boxed{4}$ Let $a;b;c$ be positive real numbers .Prove that:
$\frac {a^{3} + b^{3} + c^{3}}{a + b + c} + \frac {a^{3} + b^{3} + d^{3}}{a + b + d} + \frac {a^{3} + c^{3} + d^{3}}{a + c + d...$
Solution
By Chebyshev,we have
$\frac {a^{3} + b^{3} + c^{3}}{a + b + c} \geq \frac {a^{2} + b^{2} + c^{2}}{3}$ .
Summing up these inequalities,we get the desired result. $\Box$
$\boxed{5}$ Let $a,b,c$ be positive reals such that these satisfy $a + b + c = 3$ ,
$\frac {a}{a + 2bc} + \frac {b}{b + 2ac} + \frac {c}{c + 2ab} \leq 1$
Solution
WLOG assume $a\ge b \ge c$
Because : $(a^2 + b^2 + c^2) = \frac {1}{3}(a + b + c)(a^2 + b^2 + c^2) \ge 3abc$
Hence:
$\frac {a}{a + 2bc} + \frac {b}{b + 2ac} + \frac {c}{c + 2ab} - 1$
$= \frac {2(a - bc)}{a + 2bc} + \frac {2(b - ca)}{b + 2ac} + \frac {2(c - ab)}{c + 2ab}$
$= \frac {2(a^2 - abc)}{a^2 + 2abc} + \frac {2(b^2 - abc)}{b^2 + 2abc} + \frac {2(c^2 - abc)}{c^2 + 2abc}$
$\ge \frac {1}{3}.2(a^2 + b^2 + c^2 - 3abc)\left( \frac {1}{a^2 + 2abc} + \frac {1}{b^2 + 2abc} + \frac {1}{c^2 + 2abc}\right)$ (Using Chebyshev's inequality)
$\ge 0$ . QED $\Box$
$\boxed{6}$ (Serbia 2008)Let $a$ , $b$ , $c$ be positive real numbers such that $a + b + c = 1$ . Prove that
$\frac {1}{bc + a + \frac {1}{a}} + \frac {1}{ac + b + \frac {1}{b}} + \frac {1}{ab + c + \frac {1}{c}} \leq \frac {27}{31}.$

Solution
We have the equivalent inequality :
$\sum {\frac {9a^2 + 9abc + 9 - 31a}{a^2 + abc + 1}} \geq 0$
Now we have the comment:
Because of $a + b + c = 1$ so that $0\leq a,b,c \leq 1$
Without loss of generality, we assume $a \geq b \geq c$
But we also have the two monotonous sequences ( verify Smile

)
(a) $9( a^2 + abc + 1) - 31a \leq 9(b^2 + abc + 1) - 31b \leq 9(c^2 + abc + 1) - 31c$ ( Note that $9( a + b) \leq 31)$
(b) $\frac {1}{ a^2 + abc + 1} \leq \frac {1}{ b^2 + abc + 1} \leq \frac {1}{ c^2 + abc + 1}$
Applying Chebyshev inequality for two sequences above, we have
$\sum{\frac {9(a^2 + abc + 1) - 31a}{ a^2 + abc + 1}}$

$\geq \sum{\frac {9(a^2 + abc + 1) - 31a}{ 3}}. \sum{\frac {1}{ a^2 + abc + 1}}$
But it is easy to check by homogeneous method the following inequality:
$a^2 + b^2 + c^2 + 3abc \geq \frac {4}{9}$ $\Box$
$\boxed{7}$ Let $a,b,c > 0$ , prove that: $a^ab^bc^c\geq (abc)^{{{a + b + c}\over {3}}}$
Solution
$a^ab^bc^c\ge (abc)^{\frac {a + b + c}{3}}\Longleftrightarrow a^{3a}b^{3b}c^{3c}\ge (abc)^{a + b + c}$

$\Longleftrightarrow 3a\ln a + 3b\ln b + 3c\ln c\ge (a + b + c)(\ln a + \ln b + \ln c)$ , which follows from Chebyshev's inequality .
$\Box$
$\boxed{8}$ . For $a,b,c > 0$ , show that
$\sum_{\text{cyc}} \frac {a^3}{a^2 + ab + b^2} \ge\frac {a + b + c}{3}$

Solution
Because $\sum {\frac {{{a^3}}}{{{a^2} + ab + {b^2}}}} = \sum {\frac {{{b^3}}}{{{a^2} + ab + {b^2}}}}$
so, $LHS = \frac 12 \sum {\frac {{{a^3} + {b^3}}}{{{a^2} + ab + {b^2}}}} \ge \frac {2}{3}\sum {\frac {{{a^3} + {b^3}}}{{{a^2} + {b...$ (by Chebyshev's inequality, why? )
$\boxed{9}$ Find the maximum value of following expression
$A = \frac {x}{x^2 + yz} + \frac {y}{y^2 + zx} + \frac {z}{z^2 + xy}$ İf $x,y,z$ are positive real numbers and $x^2 + y^2 + z^2 = xyz$
Solution(Honey_S)
Rewrite the inequality as:
$\frac {x}{{x^2 + yz}} + \frac {{y^2 }}{{y^2 + zx}} + \frac {{z^2 }}{{z^2 + xy}} \le \frac {{x^2 + y^2 + z^2 }}{{2xyz}}$

$\Leftrightarrow \sum\limits_{cyc} {\left( {\frac {x}{{2yz}} - \frac {x}{{x^2 + yz}}} \right)} \ge 0$

$\Leftrightarrow \sum\limits_{cyc} {\frac {{x\left( {x^2 - yz} \right)}}{{yz\left( {x^2 + yz} \right)}}} \ge 0$
Assume $x \ge y \ge z$ then we have
$x^2 - yz \ge y^2 - zx \Leftrightarrow \left( {x - y} \right)\left( {x + y + z} \right) \ge 0$

$\frac {{x^2 }}{{x^2 + yz}} \ge \frac {{y^2 }}{{y^2 + zx}} \Leftrightarrow \frac {{z\left( {x^3 - y^3 } \right)}}{{\left( {x^2...$
Thus we have by Chebyshev inequality
$\sum\limits_{cyc} {\frac {{x^2 \left( {x^2 - yz} \right)}}{{x^2 + yz}}} \ge \frac {1}{3}\left( {\sum\limits_{cyc} {\frac {{x^...$
We finish our proof here Smile

$\Box$
$\boxed{10}$ Prove that $n^{2}(a_{1}^{3} + a_{2}^{3} + a_{3}^{3} + ... + a_{n}^{3})\geq(a_{1} + a_{2} + a_{3} + ... + a_{n})^{3}$ where $n\in\mathbb N$
Solution
Using Chebyshev's inequality
$\frac {(a_1 + a_2 + ... + a_n)}{n}\cdot \frac {(a_1 + a_2 + ... + a_n)}{n}\cdot \frac {(a_1 + a_2 + ... + a_n)}{n}$

$\leq \frac {(a_1 + a_2 + ... + a_n)}{n}\cdot \frac {a_1^2 + a_2^2 + ... + a_n^2}{n}$

$\leq \frac {a_1^3 + a_2^3 + ... + a_n^3}{n}$
. $\Box$
$\boxed{11}$ Prove that for $a,b,c > 0$ we always have:
$\frac {2a}{(b + c)^2} + \frac {2b}{(c + a)^2} + \frac {2c}{(a + b)^2}\ge\frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a...$
Solution
Assume WLOG that $a \geq b \geq c$ . Then we have $\frac {a}{b + c} \geq \frac {b}{a + c} \geq \frac {c}{a + b}$ , and $\frac {1}{b + c} \geq \frac {1}{a + c} \geq \frac {1}{a + b}$
Now by Chebyshev we have:
$LHS \geq \frac {2}{3}\left(\frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}\right)\left(\frac {1}{b + c} + \frac {1}{a ...$
$\geq \frac {2}{3}\cdot\frac {3}{2}\left(\frac {1}{b + c} + \frac {1}{a + c} + \frac {1}{a + b}\right)$
$= \frac {1}{b + c} + \frac {1}{a + c} + \frac {1}{a + b}$ $\Box$
$\boxed{12}$ . Prove that for positive reals $a,b,c$ , we have:
$\frac {4a^2}{(b + c)^3} + \frac {4b^2}{(c + a)^3} + \frac {4c^2}{(a + b)^3}\ge\frac {a}{b^2 + c^2} + \frac {b}{c^2 + a^2} + \...$

Solution
For positive reals $x,y$ we have $x^2 + y^2 \geq 2xy \Leftrightarrow 2(x^2 + y^2) \geq (x + y)^2 \Leftrightarrow (x + y)^3 \leq (x + y)2(x^2 + y^2)$ , so we have:

$\sum_{cyc}{\frac {4a^2}{(b + c)^3}} \geq \sum_{cyc}{\frac {4a^2}{2(b + c)(b^2 + c^2)}} = 2\sum_{cyc}{\frac {a}{b + c}\frac {a...$

Assume
$a \geq b \geq c$ so we have $\frac {a}{b + c} \geq \frac {b}{a + c} \geq \frac {c}{b + a}$ and $\frac {a}{b^2 + c^2} \geq \frac {b}{a^2 + c^2} \geq \frac {c}{b^2 + a^2}$

By Chebyshev inequality we have now
$2\sum_{cyclic}{\frac {a}{b + c}\frac {a}{b^2 + c^2}} \geq \frac {2}{3}\left(\frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{b...$
$\geq \frac {2}{3}\cdot\frac {3}{2}\left(\frac {a}{b^2 + c^2} + \frac {b}{a^2 + c^2} + \frac {c}{b^2 + a^2}\right) = \left(\fr...$ . QED $\Box$
$\boxed{13}$ Let $a$ ; $b$ and $c$ are positive reals numbers. Show that:
$\frac {a}{b + c + d} + \frac {b}{c + d + a} + \frac {c}{d + a + b} + \frac {d}{a + b + c} \geq \frac {4}{3}$
Solution
$(a,b,c,d)$ and $\left(\frac {1}{b + c + d},\frac {1}{a + c + d},\frac {1}{a + b + d},\frac {1}{a + b + c}\right)$ are same ordered.
Hence, $\frac {a}{b + c + d} + \frac {b}{c + d + a} + \frac {c}{d + a + b} + \frac {d}{a + b + c}$
$\geq\frac {1}{4}(a + b + c + d)\left(\frac {1}{b + c + d} + \frac {1}{a + c + d} + \frac {1}{a + b + d} + \frac {1}{a + b + c...$
$= \frac {1}{12}\sum_{cyc}(a + b + c)\sum_{cyc}\frac {1}{a + b + c}\geq\frac {1}{12}\cdot16 = \frac {4}{3}$ from Cauchy-Schwartz inequality. $\Box$
$\boxed{14}$ Prove that for arbitrary non-negative real numbers $a,b,c$ with sum $1$ we have
$\frac {x^{k + 2}}{x^{k + 1} + y^k + z^k} + \frac {y^{k + 2}}{y^{k + 1} + z^k + x^k} + \frac {z^{k + 2}}{z^{k + 1} + x^k + y^k...$
Solution
Assume that $x \ge y \ge z$ . It can be esily proved that
$\frac {x^{k + 1}}{x^{k + 1} + y^k + z^k} \ge \frac {y^{k + 1}}{y^{k + 1} + z^k + x^k} \ge \frac {z^{k + 1}}{z^{k + 1} + x^k +...$
$z^{k + 1} + x^k + y^k \ge y^{k + 1} + z^k + x^k \ge x^{k + 1} + y^k + z^k.$
Therefore applying Chebyshev's Inequality we have
$\quad\frac {x^{k + 2}}{x^{k + 1} + y^k + z^k} + \frac {y^{k + 2}}{y^{k + 1} + z^k + x^k} + \frac {z^{k + 2}}{z^{k + 1} + x^k ...$
and
$\quad\left( \frac {x^{k + 1}}{x^{k + 1} + y^k + z^k} + \frac {y^{k + 1}}{y^{k + 1} + z^k + x^k} + \frac {z^{k + 1}}{z^{k + 1}...$

The rest is trivial. Hence we are done Smile

$\Box$
$\boxed{15}$ Let $a,b,c\in\mathbb R ^ +$ ; Prove that :
$\frac {1}{2}(\frac {b + c}{a} + \frac {c + a}{b} + \frac {b + a}{c}) \geq \frac {a + b + c}{\sqrt [3]{abc}}$
Solution
From Chebyshev, since sequences $b + c;c + a;c + a$ and $\frac 1a; \frac 1b; \frac 1c$ are similarly sorted hence we have :
$\frac {1}{2}\left(\frac {b + c}{a} + \frac {c + a}{b} + \frac {b + a}{c}\right)\ge \frac {1}{3}\left(\frac {1}{a} + \frac {1}...$

$\geq \frac 13 \cdot \frac {3}{\sqrt [3]{abc}}(a + b + c) = \frac {(a + b + c)}{\sqrt [3]{abc}}$
(From AM-GM inequality).
Done. Smile

$\Box$
$\boxed{16}$ Let $x_1,x_2,...,x_n$ be distinct positive integers, then prove
$a)x{_1}^2 + x{_2}^2 + ... + x{_n}^2 \ge \frac {2n + 1}{3}(x_1 + x_2 + ... + x_n)$
Solution
Suppose that $x_1 < x_2 < ... < x_n$ ,hence $x_i \geq i$ for all $i$
Put $y_i = a_i - i$ ,the inequality becomes:
$\sum^{n}_{i = 1}b_i^2 + 2\sum^{n}_{i = 1}ib_i + \frac {n(n + 1)(2n + 1)}{6}$ $\geq \frac {2n + 1}{3}\sum^{n}_{i = 1} b_i + \frac {n(n + 1)(2n + 1)}{6}$
Note that $b_1 \leq b_2 \leq ... \leq b_n$
Then by Chebyshev inequality we have:
$2\sum^{n}_{i = 1}ib_i \geq (n + 1)\sum ^{n}_{i = 1} b_i \geq \frac {2n + 1}{3}\sum^{n}_{i = 1} b_i$
We are done. $\Box$
$\boxed{17}$ Let $a > b > c > 0$ and $abc = 1$ , Prove that:
$\frac {9}{2}\leq\sum_{(a,b,c)}\frac {1}{a^{2}(b + c)}\cdot\sum_{(a,b,c)}ab\leq3\sum_{(a,b,c)}\frac {ab}{c^{2}(a + b)}$
Solution
Let $a = \frac {1}{x},b = \frac {1}{y},c = \frac {1}{z}$ . Then using $xyz = 1$ , our inequality is equivalent to
$\frac {9}{2}\le \sum\frac {x}{y + z}\cdot \sum x\le 3\sum \frac {x^{2}}{y + z}$
For the left part, use $\sum\frac {x}{y + z}\ge \frac {3}{2}\ge \frac {9}{2(x + y + z)}$ , which is true since $x + y + z\ge 3$ .
Now we come to the right part.
Let $x\le y\le z$ so that
$x(x + z) \le y(y + z) \Leftrightarrow (x - y)(x + y + z) \ge 0,$
Hence we can apply Chebyshev:
$\sum\frac {x}{y + z}\cdot \sum x\le 3\sum \frac {x^{2}}{y + z}$
QED. Smile

$\Box$
$\boxed{18}$ Let be a,b and c non-negative reals numbers. Show that
$\frac {b + c}{a^{2}} + \frac {a + c}{b^{2}} + \frac {a + b}{c^{2}}\geq\frac {2}{a} + \frac {2}{b} + \frac {2}{c}$
Solution
rewrite the given inequality as:
$(a + b + c)\left( \frac {1}{a^{2}} + \frac {1}{b^{2}} + \frac {1}{c^{2}}\right) \geq 3\left( \frac {1}{a} + \frac {1}{b} + \f...$
after adding $\frac {1}{a}$ to $\frac {b + c}{a^{2}}$ , etc. Now the above inequality is evident by Chebyshev.
$\Box$
$\boxed{19}$ Let $x, y, z \ge 0$ be real numbers. Prove that:
$\frac {x^{3} + y^{3} + z^{3}}{3}\ge xyz + \frac {3}{4}|(x - y)(y - z)(z - x)| .$

Solution
Rewrite the inequality as:
$\frac {x^3 + y^3 + z^3 - 3xyz}{3}\geq \frac 94 |(x - y)(y - z)(z - x)|$

$\implies \left[(x - y)^2 + (y - z)^2 + (z - x)^2 \right]\left[(x + y) + (y + z) + (z + x)$

$\geq 9|(x - y)(y - z)(z - x)|$
But we have; $x + y\geq |x - y|; y + z\geq |y - z|; z + x\geq |z - x|$ . So we are only required to prove that :
$\left[(x - y)^2 + (y - z)^2 + (z - x)^2\right]\left[|x - y| + |y - z| + |z - x|\right]$

$\geq 9|(x - y)(y - z)(z - x)|$
Which is obvious from chebyshev's inequality. Smile

Equality holds iff $x = y = z = 0$ $\Box$
$\boxed{20}$ If $a,b,c > 0$ , $\sum \frac {a^{2}}{b + c}\geq \frac {\sqrt {3}}{2}\cdot \sqrt {a^{2} + b^{2} + c^{2}}\geq \frac {a + b + c}{2}$
Solution
WLOG $a\ge b\ge c$ . then from Chebyshev we have:
$\sum\frac {a^{2}}{b + c}\ge \frac {3(a^{2} + b^{2} + c^{2})}{2(a + b + c)}$
so we need to show that
$\sqrt {3(a^{2} + b^{2} + c^{2})}\ge a + b + c$
which is just the trivial inequality. Note that the right side also follows from the last inequality. $\Box$
$\boxed{21}$ Let $a,b,c$ be positive real numbers, show that
$\frac1{a\sqrt {2(a^2 + bc)}} + \frac1{b\sqrt {2(b^2 + ca)}} + \frac1{c\sqrt {2(c^2 + ab)}}\ge\frac9{2(ab + bc + ca)}$

Solution
Lemma
We claim that
$\sum_{cyc}\frac {b + c}{(a + b)(a + c)} \ge \frac {3\sqrt {3}}{2\sqrt {ab + bc + ca}}$
Hint for proof : You can use Iran 1996 Inequality. This remains as a challenge for the readers. $\Delta$
We have
$\frac {1}{a\sqrt {2(a^{2} + bc)}} = \frac {\sqrt {b + c}}{\sqrt {2a}.\sqrt {(ab + ac)(a^2 + bc)}} \ge \frac {\sqrt {2(b + c)}...$
Hence, it suffices to show that
$\sum_{cyc}\frac {\sqrt {2}(b + c)}{\sqrt {a(b + c)}.(a + b)(a + c)} \ge \frac {9}{2(ab + bc + ca)}$
Now, WLOG, we may assume that $a \ge b \ge c$ and thus,
$\left\{\begin{array} {cc} \frac {b + c}{(a + c)(a + b)} \le \frac {a + c}{(b + c)(b + a)} \le \frac {b + a}{(c + a)(c + b)} \...$
Again, using Chebyshev inequality for these and our lemma above the proof is quite obvious.
_______________________
Now since the deadline for Mathematical Reflections is over (1st Nov. 2009) hence I will post the inequality and my solutions for it.
See my attachment, it is the same document that I sent to MR, but unfortunately got no reply......
You will also get a fine demonstration of using rearrangement nicely.
Please enjoy. (Also tell me if you like my solutions)
* Link for Mathematical reflections 2009 issue 5 submissions:
http://reflections.awesomemath.org/currentissue.html
_______________________
(To be updated)
References
1. A generalization of Chebyshev's inequality. :http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244168

Mathematical Reflections.pdf

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Rearrangement Inequality

Thu Oct 08, 2009 10:55 pm, by Potla

Though underestimated and not-so-widely-used inequality; Rearrangement inequality is extremely useful in solving several Problems. Several inequalities can also be derived and proved easily from Rearrangement; such as AM-GM; Chebyshev's inequality and so on.....
Theorem
The sum $S_n = \sum_{i = 1}^n a_ib_i$ is maximal if Sequences $\{a_i\}$ and $\{b_i\}$ are similarly sorted; and minimal if sequences $\{a_i\}$ and $\{b_i\}$ are oppositely sorted.
Main Idea
If you see cyclic terms of the form $\sum a_ib_i$ then we can maximize or minimize it by rearrangement subject to given conditions. So we can write it into $S\geq \sum_{cyc}a_ib_{n - i}$ or of that structure. After forcing that structure, you can easily shift the terms that are in multiplied form and get a new structure; from which you can force identities or easier inequalities.
This idea might not be clear to the readers so I am giving us some examples of Rearrangement inequality.
Also, I will use frequently a notation :
$\begin{bmatrix} a_1 \ \ a_2 \ \ a_3 \\ b_1 \ \ b_2 \ \ b_3 \end{bmatrix} = a_1b_1 + a_2b_2 + a_3b_3$

$\begin{bmatrix} a_1 \ a_2 \ a_3 \\ b_1 \ b_2 \ b_3 \\ c_1 \ c_2 \ c_3 \end{bmatrix} = a_1b_1c_1 + a_2b_2c_2 + a_3b_3c_3$
and so on.......
Theorems Derived from Rearrangement
We can prove the AM-GM inequality for $n$ numbers using rearrangement inequality. Here I present a short proof (from Problem solving strategies; Arthur Engel):
Let $x_i > 0; c = \sqrt [n]{x_1x_2\cdots x_n}$ . Let $a_j = \frac {\prod_{i = 1}^j x_i}{c^i}$
Define also: $b_i = \frac {1}{a_i}$
We have the direct consequence $a_n = b_n = 1$
Sequences $\{a_i\}; \{b_i\}$ are oppositely sorted so that we easily have:
$\sum_{i = 1}^n a_ib_i\leq a_1b_n + a_2b_1 + \cdots a_nb_{n - 1}$
$\implies\underbrace{ 1 + 1 + \cdots1}\le \sum_{i = 1}^n \frac {x_i}{c}$
$\implies c = \sqrt [n]{x_ix_2\cdots x_n}\leq \frac1n(x_1 + x_2 + \cdots x_n)$
So we have derived AM-GM inequality. Next we will prove the Chebyshev's inequality:
Theorem 2(Chebyshev)
(a)If sequences $a_i; b_i$ are similarly sorted then
$\sum_{i = 1}^n a_ib_i\geq \frac {1}{n}(a_1 + a_2 + \cdots a_n)(b_1 + b_2 + \cdots b_n)$
And, (b)if sequences $a_i; b_i$ are oppositely sorted then we have:
$\sum_{i = 1}^n a_ib_i\leq \frac1n(a_1 + a_2 + \cdots a_n)(b_1 + b_2 + \cdots b_n)$
Proof
We have;
$a_1b_1 + a_2b_2\cdots + a_nb_n = a_1b_1 + a_2b_2\cdots + a_nb_n \\ a_1b_1 + a_2b_2\cdots + a_nb_n\geq a_1b_2 + a_2b_3 + \cdot...$
We can prove (b) also accordingly using Rearrangement Smile

Nessbitt's inequality proved by rearrangement
For $a,b,c > 0$ we have $\frac a{b + c} + \frac b{c + a} + \frac c{a + b}\geq \frac32$
We have already proved this using an elegant method by Titu's lemma. We can also prove it using AM-GM inequality, and here I show my prof using Rearrangement:
Sequences $a,b,c$ and $\frac1{b + c}; \frac {1}{c + a}; \frac1{a + b}$ are similarly sorted(why?) so from Rearrangement inequality, we have:
$\begin{bmatrix} a \ \ \ b \ \ \ c \\ \frac1{b + c} \ \frac1{c + a} \ \frac1{a + b} \end{bmatrix}\geq \begin{bmatrix} a \ \ \ ...$
So we are done.
Another problem
$a,b,c > 0;$ show that $\frac {a^2}{bc} + \frac {b^2}{ca} + \frac {c^2}{ab}\ge 3$
We have, from easy rearrangement that
$LHS\ge \frac {a^2}{ab} + \frac {b^2}{bc} + \frac {c^2}{ac} = \frac ab + \frac bc + \frac ca \geq 3$
where the last one follows from AM-GM.
______________________________________________________
Applications
$\boxed{1}$ Prove that $a^3 + b^3 + c^3\geq a^2b + b^2c + c^2a$ where $a,b,c > 0$
Indeed we have :
$a^3 + b^3 + c^3 = \begin{bmatrix} a \ \ \ b \ \ \ c \\ a^2 \ \ b^2 \ \ c^2 \end{bmatrix}\geq \begin{bmatrix} a \ \ \ b \ \ \ ...$
Q.E.D.
$\boxed{2}$ Prove that $a^4 + b^4 + c^4\geq a^2bc + b^2ca + c^2ab\forall a,b,c\geq 0$
Of course;
$a^4 + b^4 + c^4 = \begin{bmatrix} a^2 \ b^2 \ c^2 \\ a \ b \ c \\ a \ b \ c \end{bmatrix}\geq \begin{bmatrix} a^2 \ b^2 \ c^2...$
Another argument
$\boxed{3}$ Let a,b and c be positive real numbers. Prove that ${a}^a{b}^b{c}^c\geq(abc)^{\frac {a + b + c}{3}}$
Taking logarithm of both sides to the base $e$ we have:
$a\ln a + b\ln b + c\ln c\geq \left(\frac {a + b + c}{3}\right)(\ln a + \ln b + \ln c)$
But, since sequences $a,b,c$ and $\ln a, \ln b, \ln c$ are similarly sorted; and:
$\min\{a,b,c\}\leq \frac {a + b + c}{3}\leq \max\{a,b,c\}$
Hence from Rearrangement inequality, we obviously have:
$\begin{bmatrix} \ln a ;\ \ln b; \ \ln c \\ a;\ \ b;\ \ c \end{bmatrix}\geq \begin{bmatrix} \ln a\ ;\ \ \ln b\ ;\ \ \ln c \\ \...$
Hence the result. Smile

$\boxed{5}$ Let $a,b,c > 0$ Prove that; if $a^2 + b^2 + c^2 = 1$ we always have:
$\frac1{a^2} + \frac1{b^2} + \frac1{c^2}\geq 3 + \frac {2(a^3 + b^3 + c^3)}{abc}$

Solution
What happens if we apply Cauchy like $(a^2 + b^2 + c^2)\left(\frac1{a^2} + \frac1{b^2} + \frac1{c^2}\right)\geq 9$ ?
Then we must prove that $3abc\geq a^3 + b^3 + c^3$ which is impossible. So direct Cauchy gives us a weak result.
What can we do? Rearrangement or AM-GM to the rescue. We rewrite the inequality as:
$(a^2 + b^2 + c^2)\left(\frac1{a^2} + \frac1{b^2} + \frac1{c^2}\right)\geq 3 + \frac {2(a^3 + b^3 + c^3)}{abc}$
Now we just expand the LHS; so we are left to prove:
$3 + \sum_{sym} \frac {a^2}{b^2}\geq 2\sum_{cyc} \frac {a^2}{bc}+3$

$\implies \sum_{cyc} \left[\frac {a^2}{b^2} + \frac {a^2}{c^2}\right]\geq \sum_{cyc}\left(2\frac {a^2}{bc}\right)$
Which is perfectly true from Rearrangement or AM-GM inequality.
(AM-GM is applied directly here. Can you find a proof by direct rearrangement? Razz

)

$\boxed{6}$ For positive reals $a,b,c$ prove that
$\sum_{cyc}\frac {a^2 + bc}{b + c}\geq 3$
Solution(bunhiacovski)
Rewrite the given inequality as:
$\sum_{cyc}\frac {a^2}{b + c}\geq \sum_{cyc}\left[1 - \frac {bc}{b + c}\right]$
But $\sum_{cyc}\left[1 - \frac {bc}{b + c}\right]\leq \sum_{cyc}\frac {a^2}{a + b}$
And, since sequences $a^2,b^2,c^2$ and $\frac1{b + c}; \frac1{c + a}; \frac1{a + b}$ are both similarly sorted hence we finish our proof using rearrangement inequality. $\Box$
$\boxed{7}$ For $a,b,c > 0$ show that
$\sum_{cyc}\frac {a^2 + bc}{b + c}\geq a + b + c$

Solution
This is a really old and well - known problem. My proof of this shows the full power of rearrangement
There are several ways of proving this, but this is a new solution that I found, which, in my opinion, is the simplest and most elementary solution.
We rewrite the inequality as $\sum \frac {a^2}{b + c} + \sum\frac {bc}{b + c}\geq a + b + c$
Observe that sequences $\{a^2,b^2, c^2\}$ and $\left\{\frac 1{b + c}; \frac 1{c + a}; \frac 1{a + b}\right\}$ are similarly sorted.
Hence from rearrangement; $\sum \frac {a^2}{b + c}\geq \sum \frac {b^2}{b + c}$
So plugging this inequality into our required problem, we have
$LHS \geq \sum \frac {b^2 + bc}{b + c} = \sum \frac {b(b + c)}{b + c} = a + b + c$
I finish my proof here Wink

$\Box$ .

(To be Updated)
For more problems, see my next entry on Chebyshev's inequality

More inequalities of mine

Sat Oct 03, 2009 9:14 pm, by Potla

5.

6

7

8

9.

10.

11.

12.
14.

Iran 1996

Tue Aug 25, 2009 9:59 am, by Potla

[ Currently: Inequalities ]

Problem: (Iran 1996) Prove that $\forall x,y,z\in\mathbb{R}^ + ;$
$(xy + yz + zx)\left(\sum \frac {1}{(x + y)^2}\right)\geq \frac {9}{4}$

Solution 1
This is completely my own solution. . . . .

Let $x + y = a, y + z = b, z + x = c$ hence
$2x = b + c - a\implies x = \frac {b + c - a}{2}$ and other similar relations.
Hence;
$\sum yz = \frac {1}{4}\sum (2bc - a^2)$
Now assume $a\geq b\geq c$ So
$2bc - a^2\leq 2ca - b^2\leq 2ab - c^2$
So these sequences are oppositely sorted. Now we apply Chebyshev's inequality:
$\frac {4}{9}\sum yz\sum \frac {1}{(x + y^2)} = \frac {1}{3}\sum (2bc - a^2)\cdot \frac {1}{3}\sum\frac {1}{a^2}$

$\geq \frac {1}{3}\sum \left((2bc - a^2)\frac {1}{a^2}\right) = \frac {1}{3}\sum\frac {2bc}{a^2} - 1$

$\geq 2\left(\prod_{cyc} \frac {bc}{a^2}\right)^{\frac {1}{3}} - 1 = 1$
The last step follows from AM-GM inequality.
So $\frac {4}{9}\sum yz\sum\frac {1}{(x + y)^2}\geq 1 \implies \sum xy\sum \frac {1}{(x + y)^2}\geq \frac {9}{4}$
Hence proved.
(SM)
__________________________________________________
Solution 2( by Vo Quoc Ba Can):
WLOG, assume $x\ge y\ge z$ . Now we use a lemma:
$\boxed{\text{Lemma}}$
We claim that $\sum_{cyc}\frac{1}{(x+y)^2}\geq \frac{1}{4xy}+\frac{2}{(x+z)(y+z)}$
Proof: The lemma is equivalent with:
$\left(\frac{1}{x+z}-\frac{1}{y+z}\right)^2\geq \frac{(x-y)^2}{4xy(x+y)^2}$
$\Longrightarrow 4xy(x+y)^2\geq (x+z)^2(y+z)^2$
But, as $x\ge y\ge z$ so $4xy\ge 4y^2\ge (y+z)^2$
& $(x+y)^2\geq (x+z)^2$ so the lemma is proved.
$\boxed{\text{Using this lemma: }}$
The inequality
$\implies (xy+yz+zx)\left(\frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\right)\geq \frac{9}{4}$
$\implies \frac{1}{4}+\frac{z(x+y)}{4xy}+\frac{2(xy+yz+zx)}{(x+z)(y+z)}\geq \frac{9}{4}$
$\implies \frac{z(x+y)}{4xy}+2-\frac{2z^2}{(x+z)(y+z)}\geq 2$ $\left[\because 2(xy+yz+zx)=2(x+z)(y+z)-2z^2\right]$
$\implies \frac{z(x+y)}{4xy}\geq \frac{2z^2}{(x+z)(y+z)}$
$\implies (x+y)(y+z)(z+x)\geq 8xyz$ .
This easily follows from AM-GM. Hence proved.
This solution is really genial for its simplicity and brilliancy.... and also the easiest solution (in my opinion) possible. Smile

Geometry - a powerful technique

Mon Aug 17, 2009 3:02 am, by Potla

I am writing this article from the little experience & knowledge that I have gained from solving the RMO & INMO Geometry problems of our country. However, I have solved more than 7 problems in Geometry using this powerful method, which is completely based on similarity of triangles.

I am sorry that I could not post any topic due to my exams before.
Sad

A method related to similarity of triangles
We face several problems related to similarity of triangles. In some cases, similarity is used in proving theorems related to cyclic quadrilaterals. Now I come to my method.

Statement
Let ABCD be a quadrilateral whose AC & BD meet at O, such that ABO and BCO are similar. Then if $AO = a, BO = b;$ we can write $CO$ and $DO$ in terms of $a,b,k$ where $k$ is the scaling factor of the two triangles.
The triangles are similar if:
1) It is a cyclic quadrilateral;
2) If its opposite sides are parallel , ie if it is a trapezoid;

In most problems, we are given that AC and BD are perpendicular, leading to the experessions of AB, CD, .....

draw((0,50)--(0,-100));
draw((-50,0)--(100,0));
draw((-50,0)--(0,50));
draw((0,-100)--(100,0));
draw((0,50)--(100,0));
draw((...

APPLICATIONS
Here I am representing a set of problems which I solved using the method:
1. (RMO-1992; Theorem 1 in my opinion Razz

) Let ABCD be a cyclic quadrilateral such that $AC\perp BD$ ; AC meets BD at E. Prove that

where R is the radius of the circumscribing circle.
Solution:

As $ABCD$ is cyclic, $\triangle AED \sim \triangle BEC\implies \frac {EC}{EB} = \frac {ED}{EA} = k$ (say).
Letting $EA = a, EB = b\implies EC = kb, ED = ka$
Now, $AD^2 = k^2a^2 + a^2 = a^2(k^2 + 1)$ and $AB^2 = a^2 + b^2$ .
Now from $\triangle ADB$ , Sine rule gives:
$2R = \frac {AB}{\sin\angle ADB}\implies 4R^2 = \frac {AB^2}{\sin\angle ADE}$
$\therefore 4R^2 = \frac {a^2 + b^2}{\frac {a^2}{(1 + k^2)a^2}} = (1 + k^2)(a^2 + b^2)$
But $EA^2 + EB^2 + EC^2 + ED^2 = a^2 + b^2 + k^2b^2 + k^2a^2$
$= (1 + k^2)(a^2 + b^2)$
Comparing these yields the result.
_________________________________________________________________
2.Let AC and BD be chords of a circle with centre O such that the intersect at right angles inside the circle at point M. Suppose $K,L \to$ respective midpoints of $AB, CD$ respectively. Prove that $OKML$ is a parallelogram.
Solution:
(Sorry for my delay in updating)

Let $AM = a, BM = b$
Now $\triangle AMB\sim \triangle CMD$ as ABCD is a cyclic quadrilateral.
Hence $\frac {DM}{AM} = \frac {CM}{BM} = k$ (say)
So that $DM = ak; CM = bk$
AS ABM is a right triangle, $KB = \frac {1}{2}AB = \frac {1}{2}\sqrt {a^2 + b^2}$
AS CDM is a right triangle, $CL = \frac {1}{2}CD = \frac {k}{2}\sqrt {a^2 + b^2}$
Hence in $\triangle BKM$ and $\triangle CML$ ;
$\frac {BM}{MC} = \frac {BK}{LC} = \frac {1}{k};$
$\angle ACD = \angle ABD\implies \angle MCL = \angle MBK$
So that $\triangle MBK \sim \triangle CLM$
$\therefore \angle MKB - 90^{\circ} = \angle MLC - 90^{\circ}$
$\therefore\boxed{\angle MKO = \angle MLO}$
Now; triangle AMK gives:
$\cos \angle MAK = \frac {MK^2 - a^2 - \frac {1}{4}(a^2 + b^2)}{2a\frac {1}{2}\sqrt {a^2 + b^2}}$
So $MK^2 = a^2 + \frac {1}{4}(a^2 + b^2) - a\sqrt {a^2 + b^2}\cos \angle MAK$
$= a^2 + \frac {1}{4}(a^2 + b^2) - a\sqrt {a^2 + b^2}\cos \angle MDC$
$= a^2 + \frac {1}{4}(a^2 + b^2) - a\sqrt {a^2 + b^2}\cdot \frac {ak}{k\sqrt {a^2 + b^2}}$
$= \frac {1}{4}(a^2 + b^2)$
So $\boxed{MK = \frac {1}{2}\sqrt {a^2 + b^2}}$
From the problem before, we know that $R^2 = \frac {(k^2 + 1)(a^2 + b^2)}{4}$
$CL = \frac {k}{2}\sqrt {a^2 + b^2}; OC^2 = R^2 = \frac {(k^2 + 1)(a^2 + b^2)}{4}$
So $OL^2 = \frac {1}{4}(a^2 + b^2)\implies \boxed{OL = \frac {1}{2}\sqrt {a^2 + b^2}}$
So we have; $\boxed{\mathbf{MK = OL}}$

Hence from $OKML$ ; .
$ML = OL; \angle MKO = \angle MLO$
So that $OKML$ is a $||^{\text{gm}}$
QED'
________________________________________________________
3.(RMO-1997, India) In quadr. ABCD; $AB||CD$ & $AC\perp BD$ . Show that :
(a) $AD\cdot BC\ge AB\cdot CD$
(b) $AD + BC\ge AB + CD$
Solution:(by OCHA)

Let ABCD be the quadr.
$\triangle AOD\sim \triangle DOC$
So that $\frac{OC}{AO}=\frac{OD}{BO}=k$ (say)
Letting $AO=a, BO=b; \implies OC=ka, OD=kb$
We have some relations:
$AB^2=a^2+b^2$
$AD^2=a^2+k^2b^2$
$CD^2=k^2(a^2+b^2)$
$BC^2=b^2+k^2a^2$
Hence
$AD^2\cdot BC^2=a^2b^2+k^2a^4+k^2b^2+k^4a^2b^2$
$=a^2b^2(1+k^4)+(ka^2)^2+(kb^2)^2\ge 2a^2b^2k^2+(ka^2)^2+(kb^2)^2$
$=k^2(a^2+b^2)^2=AB^2\cdot CD^2$
$\therefore \boxed{AD\cdot BC\ge AB\cdot CD}$
Also; $AD^2+BC^2=AB^2+CD^2=(a^2+b^2)(1+k^2)$
$\implies AD^2+BC^2+2AD\cdot BC\geq AB^2+CD^2+2AB\cdot CD$
$\implies (AD+BC)^2\ge (AB+CD)^2$
$\boxed{\implies AD+BC\geq AB+CD}$
QED
______________________________________________________

The Floor function / Box function

Thu Jun 04, 2009 1:08 am, by Potla

Definition
We define the floor function as : $\lfloor x\rfloor =$ greatest integer less than or equal to $x$ . e.g., $\lfloor 0.786\rfloor = 0; \lfloor - 6.987\rfloor = - 7$ .

Graph of the floor function
The function is shown in the image (1st quadrant only)(see the attachment)

This function is discontinuous and indifferentiable at each and every points. Razz

Some other definitions
We define "curvy x" as: $\{ x\} = x - \lfloor x\rfloor$
We define the ceiling function as: the least integer greater than or equal to $x$ ; ie $\lceil x\rceil =$ least integer greater than or equal to $x$ . For example; $\lceil - 6.598\rceil = - 6; \lceil 8.956\rceil = 9$ .

Hence it is easy to notice that $\lceil x\rceil = 1 + \lfloor x\rfloor$ for $x\not\in\left\{\mathbb Z\cup \mathbb{R}_{<0}\right\}$

A challenge
Can you write the function $f(x)$ where $x$ is approximated to $x'$ ;i.e.:
$x = 2.569\sim 3 = x' ; x = 1.326\sim 1 = x'$ ........
Solution/DO NOT UNHIDE BEFORE YOU HAVE SOLVED IT OUT

Common concepts
$1.\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x + y\rfloor$
$2.\lfloor x\rfloor = n\Leftrightarrow n\le x\le n + 1$
$3. \left\lfloor \frac {\lfloor x\rfloor}{n}\right\rfloor = \left\lfloor \frac {x}{n}\right\rfloor$
(Proof)

Examples
$\boxed{1}$ . (alex2008) Prove that $2^{2004}|1 + \lfloor (\sqrt {13} + 3)^{2004}\rfloor$ .
(solution by Potla)
See this url for proof. (refer to post # 126, 127) Smile

$\Box$
$\boxed{2}$ For every $n\in\mathbb{N}$ ; find the largest $k\in\mathbb{N}$ , such that:
$2^k|\lfloor(3 + \sqrt {11})^{2n - 1}\rfloor$
Hint: Define $a_n = (3 + \sqrt {11})^n + (3 - \sqrt {11})^n$ . Consider different values of $a_n$ to get a recurrence relation. Then set $x = (3 + \sqrt {11})^n; y = (3 - \sqrt {11})^n$ .

$\boxed{3}$ .(alex2008 - inequation) Solve the following inequation :

$\lfloor x\rfloor^3 - (x - 1)\lfloor x\rfloor^2 \le 1 - \{x\}$
(solution by omegatheo)

$\lfloor x\rfloor^3 - (x - 1)\lfloor x\rfloor^2\e 1 - \{x\}\implies \lfloor x\rfloor ^2\left(\lfloor x\rfloor - x + 1\right)\l...$ .
$\Box$
$\boxed{4}$ (alex2008; concepts of functions needed) Find all the functions $f: \mathbb{R}\rightarrow \mathbb{R}$ such that $f(x) + f(\lfloor x\rfloor) + f(\{x\}) = 2x\ ,\ (\forall)x\in \mathbb{R}$
(solution by Farenhajt)
Putting $x = 0$ we get $f(0) = 0\quad(1)$

Putting $x = n\in\mathbb{Z}$ , we get $2f(n) + f(0) = 2n\stackrel{\mathrm{with\,(1)}}{\iff}f(n) = n\quad (2)$

Putting $x = n + \alpha$ where $n\in\mathbb{Z}$ and $0\leqslant\alpha < 1$ , we get

$f(n + \alpha) + f(n) + f(\alpha) = 2n + 2\alpha\stackrel{\mathrm{with\,(2)}}{\iff}f(n + \alpha) + f(\alpha) = n + 2\alpha\qua...$

Putting $n = 0$ in $(3)$ we get $f(\alpha) = \alpha$

Now $(3)$ becomes $f(n + \alpha) = n + \alpha$

Hence $\left(\forall x\in\mathbb{R}\right)f(x) = x$
$\Box$
$\boxed{5}$ (alex2008)Let be $a,b,c,d\in \mathbb{N}^*$ . Show that :
$\left\lfloor \frac {a^2 + b^2}{2cd + 1} \right\rfloor + \left\lfloor \frac {b^2 + c^2}{2da + 1} \right\rfloor + \left\lfloor ...$
(solution by Farenhajt)
Since $abcd\geq 1$ , we get $\left\lfloor{2\over 2abcd + 1}\right\rfloor = 0$ , hence

$a^2 + b^2 < 2cd + 1\iff a^2 + b^2\leq 2cd$
$b^2 + c^2 < 2da + 1\iff b^2 + c^2\leq 2da$
$c^2 + d^2 < 2ab + 1\iff c^2 + d^2\leq 2ab$
$d^2 + a^2 < 2bc + 1\iff d^2 + a^2\leq 2bc$

Summing the right-hand inequalities up, we get

$(a - b)^2 + (b - c)^2 + (c - d)^2 + (d - a)^2\leq 0$

and the conclusion follows.
$\Box$
$\boxed{6}$ Solve the equation $\left\lfloor \frac {x + 1}{2} \right\rfloor = \left\lfloor\frac {2x + 1}{3} \right\rfloor$ , where $\lfloor a\rfloor$ is the integer part of the real number $a$ .
Solution (makar)
Let $\ \left\lfloor\frac {x + 1}{2}}\right\rfloor = \left\lfloor{\frac {2x + 1}{3}}\right\rfloor = k,k\in\mathbb N$
$\implies k\le\frac {x + 1}{2} < k + 1$ and $\ k\le\frac {2x + 1}{3} < k + 1$
$\implies 2k - 1\le x < 2k + 1$ and $\ \frac {3k - 1}{2}\le x < \frac {3k + 2}{2}$
Now first analyze the case in which there is no solution,there are two such cases:
$2k + 1\le \frac {3k - 1}{2}$ or $\ \frac {3k + 2}{2}\le 2k - 1$
$\implies$ For $\ k\in ( - \infty, - 3]\cup [4,\infty)$ there is no solution.
$\implies$ there is a solution for $k = - 2,\ - 1,\ 0,\ 1,\ 2,\ 3$ only.
Solving for these values of $\ k$ we get
$\ \boxed{\boxed{x\in \left[ - \frac 72,\ - 3\right)\bigcup\left[ - 2,\ - 1\right)\bigcup \left[ - \frac 12,\ \frac 52\right)\...$
$\Box$
References
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273532
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273536
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273069
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=281136

MY APOLOGISES

Okay; now : PRACTICE.
EXERCISES

1. Prove that $\forall n\geq 3(n\in\mathbb{N}$ we have: $8| \left\lfloor \left(\sqrt [3]{n} + \sqrt [3]{n + 2}\right)^3\right\rfloor$

2. Prove that $\forall n\in\mathbb{N}$ we have the following identity:
$\sum_{r = 0}^n \left\lfloor x + \frac {r}{n}\right\rfloor = \lfloor nx\rfloor$
Where $x\in\mathbb{R}$ .

$\boxed{3.}$ ( important lemma)The "golden ratio" is defined as $\frac {1 + \sqrt {5}}{2}$ . Prove that : ( $n\in\mathbb{N}$ )
(a) $\lfloor \alpha(\lfloor n\alpha\rfloor - n + 1)\rfloor = n;$ or $n + 1$ .
(b) $\lfloor (n + 1)\alpha\rfloor =$ 1. $\lfloor n\alpha\rfloor + 2$ (if $\left\lfloor \alpha ( \lfloor n \alpha\rfloor - n + 1 ) \right\rfloor = n$ ) ; and 2.
$\lfloor n\alpha\rfloor + 1$ otherwise.

$\boxed{4}$ . If $a,b,c\in\mathbb{R}$ and $\lfloor na\rfloor + \lfloor nb\rfloor = \lfloor ac\rfloor \forall n\in\mathbb{N}$ we have $a\in\mathbb{Z}$ or $b\in\mathbb{Z}$ .

$\boxed5$ . (inequality; USAMO 1981) Prove that $\forall x\in\mathbb{R}^ +$ and $n\in\mathbb{N}$ we have: $\lfloor nx\rfloor = \sum_{r = 1}^n \frac {\lfloor rx\rfloor}{r}$ .

I leave you all (mainly beginners) to digest these first, and then I will post the solutions. If any of you have some solutions, please post it as a comment to this post. Smile

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Compiled problems

Blog owner: Potla
Contributors: apratimdefermat, peine, Kyoshiro, ocha, Reon

Shoutbox

Shouts

Thanks for your encouragement.

By Potla, on Thu Oct 08, 2009 11:15 pm
Nice. Keep working

By mathson, on Tue Oct 06, 2009 3:29 am
Not at all.

By Potla, on Fri Sep 04, 2009 2:35 am
Wow! You are very smart.

By Maybach, on Mon Aug 31, 2009 4:56 pm
good going

By turikachi, on Fri May 15, 2009 7:40 pm
Because of my exams..... I must do some work now.

By Potla, on Thu Apr 09, 2009 1:51 am
well, why no activity here?

By geniusbliss, on Mon Apr 06, 2009 5:46 am
Thanks for all of yours' kind( )coordination, but I cannot update this bulky( ) blog till my exams are over, so I want the contrib to continue their job (I am making peine a contrib too).......

By Potla, on Tue Mar 03, 2009 4:56 am