Inequalities Marathon

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Solution to problem 140:
Starting from the left hand side of our desired inequality we get;

$\frac {a^9}{bc} + \frac {b^9}{ca} + \frac {c^9}{ab} + \frac {2}{abc}$ = $\frac {a^{10}}{abc} + \frac {b^{10}}{abc} + \frac {c^{10}}{abc} + \frac {4}{2abc}$ .

From Cauchy Scharz we get that:

$\frac {a^{10}}{abc} + \frac {b^{10}}{abc} + \frac {c^{10}}{abc} + \frac {4}{2abc} \geq \frac {(a^5 + b^5 + c^5 + 2)^2}{5abc}$

Hence it suffices now to show that: $\frac {(a^5 + b^5 + c^5 + 2)^2}{5abc} \geq a^5 + b^5 + c^5 + 2$ or
$a^5 + b^5 + c^5 + 2 \geq 5abc$ which is true by $AM - GM$ since:

$a^5 + b^5 + c^5 + 1 + 1 \geq 5abc$ .

**Posted:** Tue Nov 03, 2009 7:21 am

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Agr_94_Math · Yang-Mills Theory Offline Joined: 17 Feb 2008 Posts: 722

Problem 141
$a,b,c$ are positive real numbers.
$\displaystyle\sum_{cyc}\sqrt {a} = 3$ . Prove that :
$8(\displaystyle\sum_{cyc}a^2) \ge 3(a + b)(b + c)(c + a)$
-Proposed by Paolo Perfetti,Universita degli studi di Tor Vergata, Itlay for Mathematical Reflections.

**Posted:** Tue Nov 03, 2009 7:39 am

Pain rinnegan · Poincare Conjecture Offline Joined: 16 Apr 2009 Posts: 176

Solution to problem 141

Problem 142 :

Let $a,b,c>0$ such that $(a+b)(b+c)(c+a)=1$ . Prove that :

$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge 3\left(1+\sqrt[3]{\frac{1}{abc}}\right)\ge 9$

**Posted:** Tue Nov 03, 2009 12:32 pm

Abdek

another solution to problem 141:

Using Holder inequality we have :

$\sum_{cyc}a^2.\sum_{cyc}\sqrt {a}.\sum_{cyc}\sqrt {a} \ge (a + b + c)^3$

which is equivalent to :

$\sum_{cyc}a^2 \ge \frac {(a + b + c)^3}{9}$

and By AM GM we have : $8(a + b + c)^3 = (a + b + b + c + c + a)^3 \ge 27(a + b)(b + c)(c + a)$

and hence $\text{LHS} \ge \frac {27}{9}\prod(a + b)$

which is equivalent to our problem:)

**Posted:** Tue Nov 03, 2009 12:37 pm

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Mharchi Abdelmalek

Abdek

Solution to problem :142

Let $abc = t$ we have $(a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - t$

and it follows that $1 + t \ge 9t$ acording AM GM inequality which is equivalent to $t\le \frac {1}{8}$

and hence $3(1 + \sqrt [3]{\frac {1}{abc}}) \ge 9$

on the other hand we have :
$(a + b + c)(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}) \ge 3(1 + \sqrt [3]{\frac {1}{abc}})$
$\Leftrightarrow (a + b + c)(ab + bc + ca) \ge 3(t + t^{\frac {2}{3}})$
$\Leftrightarrow 1 + t \ge 3(t + t^{\frac {2}{3}})$
$\Leftrightarrow 1 \ge 2t + 3.t^{\frac {2}{3}}$

which is true since $t \le \frac {1}{8}$

**Posted:** Tue Nov 03, 2009 12:52 pm

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Mharchi Abdelmalek

Tomekk

Hello,

I menaged to prove only second part of this inequality,
which is alot easier than the first part.

Part of solution for 142

EDIT: I apologise Abdek, havent seen your post. Blush

**Posted:** Tue Nov 03, 2009 1:04 pm

socrates · Yang-Mills Theory Offline Joined: 29 Jun 2005 Posts: 740

Next one...

Problem 143

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

Smile

To those responding, don't do this problem yet. Problem 142 has two sides with only one proven, so it needs to be completed first. - mod

**Posted:** Tue Nov 03, 2009 1:42 pm

socrates · Yang-Mills Theory Offline Joined: 29 Jun 2005 Posts: 740

**Posted:** Wed Nov 04, 2009 10:25 am

Abdek

Solution to problem 143:

Using AM_GM inequality we have :

$\text{LHS} \ge \sum_{cyc}\frac {2ab}{c}$

and Thus it's sufficient to show that :

$\sum_{cyc}\frac {ab}{c} \ge ab + bc + ca$

By Cauchy shwarz inequality we have:

$\left(\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}\right)(abc + abc + abc) \ge (ab + bc + ca)^2$

and hence we need only prove that:

$ab + bc + ca \ge 3abc$ which is equivalent to $\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)^2 \ge 9$

and by AM GM we have :

$\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)^2 \ge 3\left(\frac {1}{ab} + \frac {1}{bc} + \frac {1}{ca} + \frac {1...$

On the other hand we can easily prove that $abc \ge 1$

so $3\left(\frac {1}{ab} + \frac {1}{bc} + \frac {1}{ca} + \frac {1}{abc}\right) - \frac {3}{abc} = 3.4 - \frac {3}{abc} \ge 9$

**Posted:** Wed Nov 04, 2009 1:30 pm

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Mharchi Abdelmalek

Abdek

Problem 144:

For $a,b,c>0$ prove that :

$abc(ab+bc+ca) \ge (a^2+b^2+c^2)(a+b-c)(b+c-a)(c+a-b)$

**Posted:** Wed Nov 04, 2009 1:41 pm

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Mharchi Abdelmalek

peine · Riemann Hypothesis Offline Joined: 07 Aug 2008 Posts: 280

Solution to problem 144:
Problem 145:
Prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0$ for all $a,b,c$ lengthes of side of triangle.
P.S: sorry if it's easy,

**Posted:** Thu Nov 05, 2009 5:42 am

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geniusbliss

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Abdek

**Posted:** Thu Nov 05, 2009 7:12 am

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Mharchi Abdelmalek

hasan4444 · Riemann Hypothesis Offline Joined: 28 Nov 2008 Posts: 481

Edited: Sorry I didn't notice that I posted the hard general case before.

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geniusbliss

**Posted:** Thu Nov 05, 2009 9:39 am

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geniusbliss

**Posted:** Thu Nov 05, 2009 9:49 am

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sahilsharma94 · Poincare Conjecture Offline Joined: 07 Aug 2009 Posts: 121

sorry guys i dont know latex..

solution to that summation [a/(b+c) + {ab + bc+ ca}/a^2 + b^2 + c^2 ] problem...
(a^2 + b^2)/2 >=ab

using this thrice for ab,bc,ca and adding we get a^2 + b^2 + c^2 >= ab + bc + ca
=> 1 >= (ab + bc + ca)/ a^2 + b^2 + c^2
so the prob reduces to prooving that a/(b+c) + b/(c+a) + c/(b+a) >= 3/2..which is true(it is nesbitt's inequality)