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Algebra

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3 Feb 2009 20:07:03 IST

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Inequality

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Prove that for all real x, we have Sorry, last term numerator is x2-z2 (This after I thought i had proof-read it twice )

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Rahul Duggal

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Joined: 10 May 2008

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3 Feb 2009 22:40:21 IST

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Hari Shankar

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4 Feb 2009 13:36:21 IST

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Thats a nice substitution Rahul!

To take full advantage of that, you can write that the inequality is now equivalent to proving that

$\frac{b-a}{a} + \frac{c-b}{b} + \frac{a-c}{c} \ge 0$ or $\frac{b}{a} + \frac{c}{b} + \frac{a}{c} \ge 3$ which is true by AM-GM

I wrote the inequality this way:

$\frac{x^2}{2x^2+1} + \frac{y^2}{2y^2+1} + \frac{z^2}{2z^2+1} \le \frac{y^2}{2x^2+1} + \frac{z^2}{2y^2+1} + \frac{x^2}{2z^2+1}$

and this is true as for any ordering of (x,y,z), the LHS is the least sum possible among all permutations according to Rearrangement Inequality.

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