Nice Inequality

by utkarshgupta, Oct 4, 2014, 9:05 AM

ISLs :D
Problem (ISL 2006 A5):

If $a,b,c$ are the sides of a triangle, prove that
\[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]

Solution :
My solution is easy

Since $a,b,c$ are the sides of a triangle, we can set
$a=(q+r)^2$, $b=(p+r)^2$, $c=(p+q)^2$

Thus the inequality can be rewritten as
$\sum_{cyclic} \frac{\sqrt{2p^2+2pr+2pq-2qr}}{p} \le 6$

Thus is it is sufficient to show that
$\sum_{cyclic} \frac{2p^2+2pr +2pq -2qr}{p^2} \le 12$
$\iff \sum_{cyclic} \frac{pr+pq-qr}{p^2} \le 3$
$\iff \sum_{cyclic} q^2r^2(pr+pq-qr) \le 3p^2q^2r^2$

This is third degree Schur's inequality for $pq,qr,rp$

Hence proved

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2 Comments

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clever solution :)

by Ashutoshmaths, Oct 6, 2014, 3:28 AM

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Thanks :D
It took me nearly an hour :P

by utkarshgupta, Oct 7, 2014, 11:31 AM

Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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  • Nice blog ! Your isogonality lemma is really powerful !

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  • alas,this is ded

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  • Thanks for the nice blog.

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  • I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it :D

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  • Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain :D

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  • hey utkarsh jee is over now ... continue your elementary blog pleaseeeeeee!

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  • INTERSTING BLOG

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  • I have no plans for this blog right now....
    No time here people !
    But lets see....
    I may try some combinatorics :P

    by utkarshgupta, Feb 15, 2017, 12:47 PM

  • Thanks for the nice blog!

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  • Revive it!!!
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