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Inequalities Marathon
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hasan4444
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#1
 Inequalities Marathon
Pre-Olympiad Level
Hello Everyone,

This is a try to make a nice inequalities marathon in the Pre-Olympiad level

Please if you write any problem don't forget to indicate its number and if you write a solution please indicate for what problem also to prevent the confusion that happens in some marathons.

Please show your solution don't just write by AM-GM then Cauchy-Schwarz and we are done.

OK finishing the talk now we go:

Problem 1: For any positive real numbers a,b,c show that the following inequality holds \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b}+\frac{a+b}{a+c}+\frac{b+c}{b+a}

PostPosted: Mon Sep 07, 2009 1:50 pm  Back to top 
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alex2008
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#2
Solution to problem 1
Ok. After not so many computations i got that:
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} - \frac {a + b}{c + a} - \frac {b + c}{a + b} - \frac {c + a}{b + c}

= \frac {abc}{(a + b)(b + c)(c + a)}\left(\frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2} - \frac {a}{b} - \frac {b}{c...

+ \frac {abc}{(a + b)(b + c)(c + a)}\left(\frac {ab}{c^2} + \frac {bc}{a^2} + \frac {ca}{b^2} - 3\right)
So in order to prove the above inequality we need to prove \frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2}\ge \frac {a}{b} + \frac {b}{c} + \frac {c}{a} and \frac {ab}{c^2} + \frac {bc}{a^2} + \frac {ca}{b^2}\ge 3

The second inequality is obvious by AM-GM , and the for the first we have:
\left(\frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {c^2}{a^2}\right)^2\ge 3\left(\frac {a^2}{b^2} + \frac {b^2}{c^2} + \frac {...
where i used AM-GM and the inequality 3(x^2 + y^2 + z^2)\ge (x + y + z)^2 for x = \frac {a}{b}\ ,\ y = \frac {b}{c}\ ,\ z = \frac {c}{a}

So the inequality is proved.


Problem 2:Let a,b,c,d > 0 such that a\le b\le c\le d and abcd = 1 . Then show that:
(a + 1)(d + 1)\ge 3 + \ds\frac {3}{4d^3}
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PostPosted: Mon Sep 07, 2009 2:14 pm  Back to top 
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#3
Another solution to 1

Substitute \frac {a}{b} = x,\frac {b}{c} = y,\frac {c}{a} = z.So xyz = 1.The inequality after substitution becomes
x^{2}z + y^{2}x + z^{2}x + x^{2} + y^{2} + z^{2}\ge x + y + z + 3

x^{2}z + y^{2}x + z^{2}x\ge3.So now it is left to prove that x^{2} + y^{2} + z^{2}\ge x + y + z which is easy.


EDIT:alex2008 where is b and c in your inequality.
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PostPosted: Tue Sep 08, 2009 4:58 am  Back to top 
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alex2008
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#4
Maths Mechanic wrote:
Another solution to 1

Substitute \frac {a}{b} = x,\frac {b}{c} = y,\frac {c}{a} = z.So xyz = 1.The inequality after substitution becomes
x^{2}z + y^{2}x + z^{2}x + x^{2} + y^{2} + z^{2}\ge x + y + z + 3

x^{2}z + y^{2}x + z^{2}x\ge3.So now it is left to prove that x^{2} + y^{2} + z^{2}\ge x + y + z which is easy.


EDIT:alex2008 where is b and c in your inequality.


There is no b,c in my inequality . The inequality is correct . You can try to find a counterexample but you'll only lose your time . Anyway the inequality is easy . b,c are not so important , they are more for confusing .

@Maths Mechanic: Are you Raghav Grover?
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PostPosted: Tue Sep 08, 2009 6:01 am  Back to top 
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Mateescu Constantin
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#5
Solution to problem 2


From the condition a\leq b\leq c\leq d we get that a\geq\frac {1}{d^3}.

\implies (a + 1)(d + 1)\geq \left (\frac {1}{d^3} + 1\right )(d + 1)

Now let's prove that \left (1+\frac{1}{d^3}\right )(d+1)\geq 3 + \frac {3}{4d^3} .

This is equivalent with: (d^3 + 1)(d + 1)\geq 3d^3+\frac 34

\Longleftrightarrow \left [d(d - 1)\right ]^2 - [d(d - 1)]+1 \ge \frac 34 \Longleftrightarrow \left [d(d - 1)-\frac {1}{2}\r...

Equality holds for a = \frac {1}{d^3} and d(d - 1) - \frac {1}{2} = 0\Longleftrightarrow d = \frac {1 + \sqrt {3}}{2}



PostPosted: Tue Sep 08, 2009 6:18 am  Back to top 
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#6
alex2008 wrote:

@Maths Mechanic: Are you Raghav Grover?
.

Yes Very Happy
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PostPosted: Tue Sep 08, 2009 7:04 am  Back to top 
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hasan4444
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#7
Problem 3
Darij Grinberg
Problem 3: If a,b,c are three positive real numbers, then
\frac {a}{\left(b + c\right)^2} + \frac {b}{\left(c + a\right)^2} + \frac {c}{\left(a + b\right)^2} \geq \frac {9}{4\left(a +...
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PostPosted: Tue Sep 08, 2009 7:12 am  Back to top 
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Dimitris X
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#8
solution on problem 3

\sum\frac {a^2}{ab^2 + ac^2 + 2abc} \ge \frac {(a + b + c)^2}{\sum _{sym}a^2b + 6abc}
So we only have to prove that:
4(a + b + c)^3 \ge 9 \sum_{sym}a^2b + 54abc \Longleftrightarrow 4(a^3 + b^3 + c^3) + 12\sum_{sym}a^2b + 24abc \ge 9\sum_{sym}....

But \sum_{sym} a^2b \ge 6abc.And a^3 + b^3 + c^3 \ge 3abc.
So 4(a^3 + b^3 + c^3) + 3\sum_{sym}a^2b \ge 30 abc


PROBLEM 4
For a,b,c \ge 0 and a + b + c = 1 prove that 7(ab + bc + ca) \le 2 + 9abc
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PostPosted: Tue Sep 08, 2009 7:30 am  Back to top 
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alex2008
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#9
Solution to problem 4
Homogenize to
2(a + b + c)^3 + 9abc\geq 7(ab + bc + ca)(a + b + c)
Expanding it becomes :
\sum_{sym} a^3 + 6\sum_{sym} a^2b + 21abc\geq 7\sum_{sym} a^2b + 21abc
So we just need to show:
\sum_{sym} a^3\geq \sum_{sym} a^2b
which is obvious by

a^3 + a^3 + b^3\geq 3a^2b and similars.


Problem 5:Let x,y,z\in \mathbb{R}_ + . Pove that :
\sqrt {x(y + 1)} + \sqrt {y(z + 1)} + \sqrt {z(x + 1)}\le \frac {3}{2}\sqrt {(x + 1)(y + 1)(z + 1)}
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PostPosted: Tue Sep 08, 2009 7:40 am  Back to top 
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modularmarc101
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#10
Sorry to interrup the flow of the marathon guys, but where did you learn about inequalities at from this level? Cuz I finished the Inequalities chapters in Intermediate Algebra and AoPS Vol. 2 but I guess this is the next level Very Happy
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PostPosted: Tue Sep 08, 2009 2:51 pm  Back to top 
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varunrocks
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#11
I guess these would be old IMO or MO's or TST's. Because they do not give any inequalities in IMO because have become so good at them!
x+1=a/b,y+1=b/c,z+1=c/a.
sqrt(a-b/c)+sqrt(b-c/a)+sqrt(c-a/b)<=3/2
sqrt(a-b/c)<=1/2
a-b/c<=1/4
4(a-b)<=c
4(b-c)<=a
4(c-a)<=b
Adding all equations
0<=a+b+c

PostPosted: Tue Sep 08, 2009 6:16 pm  Back to top 
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Rofler
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#12
Your solution assumes (x + 1)(y + 1)(z + 1) = 1, a very strange assumption to make, and cannot be done since the inequality is not homogenous.

PostPosted: Tue Sep 08, 2009 6:38 pm  Back to top 
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enndb0x
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#13
solution to problem 5

Dividing with the square root on the RHS we have :

\sqrt{\frac{x}{(x+1)(z+1)}} +\sqrt{\frac{y}{(x+1)(y+1)}} +\sqrt{\frac{z}{(y+1)(z+1)}} \leq \frac{3}{2}

By AM-GM

\sqrt{\frac{x}{(x+1)(z+1)}} \leq \frac{1}{2} \left( \frac{x}{x+1} +\frac{1}{y+1} \right)

\sqrt{\frac{y}{(x+1)(y+1)}} \leq \frac{1}{2} \left(\frac{y}{y+1} +\frac{1}{x+1} \right)

\sqrt{\frac{z}{(y+1)(z+1)}} \leq \frac{1}{2}\left(\frac{z}{z+1} +\frac{1}{y+1} \right)

Summing we obtain

LHS \leq \frac{1}{2} \left( \left(\frac{x}{x+1} +\frac{1}{x+1} \right) +\left(\frac{y}{y+1} +\frac{1}{y+1} \right) +\left(\fr...



Problem 6 . Let a,b,c be positive numbers , then prove that
\frac{1}{a} +\frac{1}{b} +\frac{1}{c} \geq \frac{4a}{2a^2 +b^2 +c^2} +\frac{4b}{a^2 +2b^2 +c^2} +\frac{4c}{a^2 +b^2 +2c^2}

PostPosted: Wed Sep 09, 2009 2:56 am  Back to top 
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Mateescu Constantin
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#14
Solution to problem 6


By AM-GM we have 2a^2+b^2+c^2\ge 4a\sqrt{bc} .

\implies \frac{4a}{2a^2+b^2+c^2}\le \frac{4a}{4a\sqrt{bc}}=\frac{1}{\sqrt{bc}} .

Addind the similar inequalities \implies RHS\le \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ (1) .

Using Cauchy-Schwarz we have \left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)^2\le \left(\frac 1a+\frac 1b+\frac 1c\right)^2

so \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\le\frac 1a+\frac 1b+\frac 1c\ (2) .

From (1),(2) we obtain the desired result .



Problem 7

Let a,b,c,d,e be non-negative real numbers such that a+b+c+d+e=5 . Prove that:

abc+bcd+cde+dea+eab\le 5 .

PostPosted: Wed Sep 09, 2009 3:12 am  Back to top 
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#15
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for problem 7:
apply AM GM 3 at a time to (a,b,c,d,e).on adding we get it
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PostPosted: Wed Sep 09, 2009 3:31 am  Back to top 
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#16
No that wouldn't work. You would get abc + bcd + cde + abd + abe \leq \frac{1}{27}[ (a+b+c)^3 + (b+c+d)^3 + (c + d + e)^3 + (a + b + d)^3 + (a + b + e)^3 ]
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PostPosted: Wed Sep 09, 2009 3:46 am  Back to top 
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alex2008
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#17
Solution to problem 7
Assume e\le \min\{a,b,c,d\} . Then AM-GM gives :

e(c+a)(b+d)+bc(a+d-e)\le \frac{e(5-e)^2}{4}+\frac{(5-2e)^2}{27}\le 5

the last one being equivalent with:

(e-1)^2(e+8)\ge 0


Problem 8 Let a,b,c be real numbers such that 0<a\le b\le c . Prove that:

(a+b)(c+a)^2\ge 6abc
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PostPosted: Wed Sep 09, 2009 4:49 am  Back to top 
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hasan4444
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#18
Re: Inequalities Marathon
Pre-Olympiad Level
@aadil
Don't forget that rule please
hasan4444 wrote:

Please show your solution don't just write by AM-GM then Cauchy-Schwarz and we are done.

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PostPosted: Wed Sep 09, 2009 5:40 am  Back to top 
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#19
Solution to 8

Substitute b=ax,c=by=axy
Then we are left to prove that 1+2xy+x+x^{3}y^{2}\ge3x^{2}y
which is true by A.M>G.M..


Problem 9
Prove for positive reals
\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge 2
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PostPosted: Wed Sep 09, 2009 11:55 pm  Back to top 
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Dimitris X
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#20
solution to 9

From andreescu LHS \ge \frac{(a+b+c+d)^2}{ \sum_{sym} ab+(ac+bd)}
So we only need to prove that:
(a+b+c+d)^2\ge 2\sum_{sym} ab+2(ac+bd) \Longleftrightarrow (a-c)^2+(b-d)^2 \ge 0....


PROBLEM 10
Let a,b,c,d be REAL numbers such that a^2+b^2+c^2+d^2=4.
Prove that a^3+b^3+c^3+d^3 \le 8
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