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Inequalities Marathon
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Dimitris X
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#321
solution to problem 146

Setting a = x + y,b = y + z,c = z + x, the inequality can be rewritten:

\sum_{cyc}\frac {x + y}{2z + x + y} + \frac {\sum x^2 + 3\sum xy}{2(\sum x^2 + \sum xy)} \le \frac {5}{2} \Longleftrightarrow...
which is clearly true from cauchy-swharz inequality Wink

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PostPosted: Thu Nov 05, 2009 1:58 pm  Back to top 
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peine
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#322
the problem 146 is extremetly nice, and Dimitrix's solution is also nice Smile , this is another solution,
Solution to problem146:

\sum \frac{a}{b+c}+\frac{ab+bc+ca}{a^2+b^2+c^2} \leq \frac{5}{2}
\Leftrightarrow \frac{1}{2}\left(\sum \frac{a}{b+c}+\frac{c}{a+b} -1\right)+\frac{ab+bc+ca-a^2-b^2-c^2}{a^2+b^2+c^2}\leq0
\Leftrightarrow \sum\frac{(a-b)^2}{(a+c)(b+c)}-\frac{\sum(a-b)^2}{a^2+b^2+c^2}
\Leftrightarrow S=S_a(b-c)^2+S_b(a-c)^2+S_c(a-b)^2\geq0
the inequality is symetric, then we can assume that a\geq b\geq c, then:S_a\geq0 and \frac{a-b}{a-c}\geq \frac{c}{b}
then S\geq\left(\frac{a-c}{b}\right)^2\left(b^2S_b+c^2S_c\right)
and we have b^2S_b+c^2S_c= bc(b-c)^2+a(b+c-a)(b^2+c^2)\geq 0
From where we deduct that S\geq 0 equality holds when a=b=c

I think that Dimitrix must post a new Problem.
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PostPosted: Thu Nov 05, 2009 2:24 pm  Back to top 
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socrates
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#323
Problem 147

Let a,b,c be positive real numbers such that \displaystyle{\frac {b^2 + c^2}{a} + \frac {c^2 + a^2}{b} + \frac {b^2 + a^2}{c}\leq 2(ab + bc + ca)}.
Prove that 4abc \geq a + b + c + 1.

Smile

PostPosted: Thu Nov 05, 2009 4:00 pm  Back to top 
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Abdek
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#324
Sorry EDITED Embarassed
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Last edited by Abdek on Fri Nov 06, 2009 8:22 am; edited 1 time in total 
PostPosted: Fri Nov 06, 2009 5:09 am  Back to top 
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peine
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#325
Abdek, the inequality is true if we substitu the variables with the values that you gave, I find this problem very nice,
Solution to problem 147:

the condition is equivalent as,
\sum ab(a^2+b^2) \leq 2abc(\sum ab) using this we find,
1-> \frac{2}{3}(ab+bc+ca)^2\leq 2\sum a^2b^2 \leq \sum ab(a^2+b^2) \leq 2abc(\sum ab)
\Rightarrow ab+bc+ca\leq 3abc
2-> abc(a+b+c)\leq \sum a^2b^2 \leq abc(ab+bc+ca)
\Rightarrow ab+bc+ca \geq a+b+c
3-> by AM-GM 3abc \geq ab+bc+ca \geq 3\sqrt[3]{a^2b^2c^2}
\Rightarrow abc \geq 1
then from these three result we get,
4abc \geq ab+bc+ca+abc \geq a+b+c+1
equality holds when a=b=c=1
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PostPosted: Fri Nov 06, 2009 7:28 am  Back to top 
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socrates
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#326
Problem 148

Find the minimum possible value of the constant k, such that for any nonegative real numbers a,b,c, not all zero,

satisfying a+b+c=ab+bc+ca, the following inequality holds (a+b+c) \left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}-k \right) \leq k.

Smile

PostPosted: Fri Nov 06, 2009 10:13 am  Back to top 
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Pain rinnegan
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#327
socrates wrote:
Problem 148

Find the minimum possible value of the constant k, such that for any nonegative real numbers a,b,c, not all zero,

satisfying a + b + c = ab + bc + ca, the following inequality holds (a + b + c) \left( \frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a} - k \right) \leq k.

Smile


See here .
Last edited by Pain rinnegan on Fri Nov 06, 2009 11:15 am; edited 1 time in total 
PostPosted: Fri Nov 06, 2009 10:40 am  Back to top 
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socrates
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#328
No, it's not the same problem. See it again. Smile

PostPosted: Fri Nov 06, 2009 10:56 am  Back to top 
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#329
socrates wrote:
No, it's not the same problem. See it again. Smile


The same or not the same , i meant that arqday's solution is the same as for the inequality you posted . Rewrite the inequality as :
\frac {(a + b + c)(ab + bc + ca)}{(a + b)(b + c)(c + a)}\le k
So k_{min} = \frac {9}{8}

Problem 149. Let a,b,c\ge 0 such that a^2 + b^2 + c^2 = 3 . Show that :
4 + \frac {5}{abc}\ge a + b + c + 1 + \frac {5}{abc}\ge 3\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)\ge 9

PostPosted: Fri Nov 06, 2009 11:13 am  Back to top 
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#330
Pain rinnegan wrote:
Problem 149. Let a,b,c\ge 0 such that a^2 + b^2 + c^2 = 3 . Show that :
4 + \frac {5}{abc}\ge a + b + c + 1 + \frac {5}{abc}\ge 3\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)\ge 9

Solution to Problem 149
First Part
We prove that 4+\frac{5}{abc}\ge a+b+c+1+\frac{5}{abc}\implies 3\ge a+b+c.

Since a^2+b^2+c^2=3, it suffices to prove that a^2-a+b^2-b+c^2-c\ge 0.

Consider the function f(x)=x-\sqrt{x}. It can be easily shown that it is convex in (0,3).

Hence, f(a^2)+f(b^2)+f(c^2)\ge 3f\left(\frac{a^2+b^2+c^2}{3}\right)=3f(1)=0, as desired.


Second Part
No solution. Sad


Third Part
We prove that 3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge 9\implies ab+bc+ca\ge 3abc.

Since \frac{ab+bc+ca}{3}\ge \sqrt[3]{a^2b^2c^2}, it suffices to prove that 3\sqrt[3]{a^2b^2c^2}\ge 3abc\implies 1\ge abc.

However, from 1=\frac{a^2+b^2+c^2}{3}\ge \sqrt[3]{a^2b^2c^2} we have 1\ge abc, and we are done.


I have been trying to prove the second part of the inequality chain for hours, but I couldn't...

Any ideas for the second part?
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PostPosted: Sun Nov 08, 2009 1:38 pm  Back to top 
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#331
Pain rinnegan wrote:

Problem 149. Let a,b,c\ge 0 such that a^2 + b^2 + c^2 = 3 . Show that :
4 + \frac {5}{abc}\ge a + b + c + 1 + \frac {5}{abc}\ge 3\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)\ge 9


Try a=\frac{1}{\sqrt{3}}, b=\frac{2}{\sqrt{3}}, c=\frac{2}{\sqrt{3}} for the 2nd inequality. Then we get

\frac{5\sqrt{3}}{3} + 1 + \frac{15\sqrt{3}}{4} \ge 6\sqrt{3}
\Leftrightarrow 1 \ge \frac{7\sqrt{3}}{12}
\Leftrightarrow 144 \ge 147

which is false.
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PostPosted: Tue Nov 10, 2009 5:26 pm  Back to top 
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#332
I tried the second inequality too.I couldnt find any counterexample though.Could you provide a counterexample for distinct positive reals
a,b,c?
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PostPosted: Wed Nov 11, 2009 5:38 am  Back to top 
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Pain rinnegan
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#333
not_trig wrote:
Pain rinnegan wrote:

Problem 149. Let a,b,c\ge 0 such that a^2 + b^2 + c^2 = 3 . Show that :
4 + \frac {5}{abc}\ge a + b + c + 1 + \frac {5}{abc}\ge 3\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)\ge 9


Try a = \frac {1}{\sqrt {3}}, b = \frac {2}{\sqrt {3}}, c = \frac {2}{\sqrt {3}} for the 2nd inequality. Then we get
\frac {5\sqrt {3}}{3} + 1 + \frac {15\sqrt {3}}{4} \ge 6\sqrt {3}

\Leftrightarrow 1 \ge \frac {7\sqrt {3}}{12}

\Leftrightarrow 144 \ge 147
which is false.


Ok , i'm sorry . I'll change it with the following :

Problem 149. Let a,b,c\ge 0 . Find the maximum value of P , where :

P=2(a+b+c)+abc-a^3-b^3-c^3

PostPosted: Wed Nov 11, 2009 10:33 am  Back to top 
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#334
Solution to problem 149

P = 2(a + b + c) - (a^{3} + b^{3} + c^{3}) + abc\leq S = 2(a + b + c) - \frac {(a + b + c)^{3}}{3} + \frac {(a + b + c)^{3}}{...
We consider the function :
f(x) = 2x - \frac {8x^{3}}{27}
with a+b+c=x
f'(x) = 2 - \frac {24x^{2}}{27}
We obtain maximum for :
f(\frac {3}{2}) = 2
So :
P\leq S\leq2

Last edited by b.s.o on Wed Nov 11, 2009 4:42 pm; edited 2 times in total 
PostPosted: Wed Nov 11, 2009 4:08 pm  Back to top 
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#335
b.s.o wrote:
Solution to problem 149

P = 2(a + b + c) - (a^{3} + b^{3} + c^{3}) + abc\leq S = 2(a + b + c) - \frac {(a + b + c)^{3}}{3} + \frac {(a + b + c)^{3}}{...
We consider the function :
f(x) = 2x - \frac {8x^{3}}{27}
with x+y+z=a
f'(x) = 2 - \frac {24x^{2}}{27}
We obtain maximum for :
f(\frac {1}{3}) = \frac {1}{2}
So :
P\leq S\leq \frac {1}{2}

a=b=c=1\implies P=2(1+1+1)+1-1-1-1=4 Wink
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PostPosted: Wed Nov 11, 2009 4:14 pm  Back to top 
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ocha
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#336
easily fixed

P \le 2(a+b+c) - \frac{(a+b+c)^3}{9} + \frac{(a+b+c)^3}{27} = 2(a+b+c) - \frac{2(a+b+c)^3}{27}

AM-GM
\frac{2(a+b+c)^3}{27} + 2 + 2 \ge \sqrt[3]{\frac{2^3(a+b+c)^3}{27}} = 2(a+b+c)

So the max is 4

PostPosted: Wed Nov 11, 2009 5:27 pm  Back to top 
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#337
Problem 150
(Indian Regional Mathematical Olympiad 2008)
Let a,b\in\mathbb R satisfy the condition that the roots of the cubic equation
ax^3 - x^2 + bx - 1 = 0 are all positive reals. Prove that a,b also satisfy:
(a) 0<3ab\leq 1
(b) b\geq \sqrt {3}
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PostPosted: Thu Nov 12, 2009 12:46 am  Back to top 
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TRAN THAI HUNG
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#338
Using Viete theorem
Let x,y,z is the roots of the equation. We have
x+y+z=1/a
xy+yz+xz=b/a
xyz=1/a

Then it easily to prove that a,b>0
ab = \frac{{xy + yz + xz}}{{(x + y + z)^2 }} \le \frac{1}{3}
b^2 = \frac{{(xy + yz + xz)^2 }}{{xyz(x + y + z)}} \ge 3
Then(a) 0<3ab\leq 1
and (b) b\geq\sqrt{3}
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PostPosted: Thu Nov 12, 2009 3:15 am  Back to top 
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Dimitris X
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#339
solution to problem 150

Let us denote x,y,z the real roots of the cubic equation.

From vieta's relations we can easily found that:

x + y + z = \frac {1}{a}.

xy + yz + zx = \frac {b}{a}.

xyz = \frac {1}{c}.

For the first it suffices to prove that
ba \le \frac {1}{3} \Longleftrightarrow \frac {b}{a} \cdot a^2 \le \frac {1}{3} \Longleftrightarrow \frac {xy + yz + zx}{(x +... which is clearly true.....

Now for the second:

b \ge \sqrt {3} \Longleftrightarrow \frac {xy + yz + zx}{x + y + z} \ge \sqrt {3} \Longleftrightarrow (xy + yz + zx)^2 \ge 3(....
Which is true....

Because xyz = x + y + z = \frac {1}{a}.


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PostPosted: Thu Nov 12, 2009 3:40 am  Back to top 
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#340
Let 0<a,b,c<1 such that a+b+c=2. Prove the following inequality:

1<ab+bc+ca-abc \leq \frac{28}{27}

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PostPosted: Thu Nov 12, 2009 5:26 am  Back to top 
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