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Differential Calculus

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27 Nov 2008 04:14:38 IST

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An inequality: Everyone's invited

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Let $p$ and $q$ be positive reals such that $p+q=1$ . Prove that for all $x\in \mathbb{R}$ ,

$pe^{x/p}+qe^{-x/q}\leq e^{x^2/8p^2q^2}$

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Anant Kumar

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4 Dec 2008 16:22:16 IST

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nobody trying?

akki ~~ unlucky forever ~~

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16 Jan 2009 12:01:12 IST

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ok sir here's my solution :

without the loss of generallity let

now consider two sets, and $(e^ rac{x}{p},e^ rac{-x}{q})$

now , by chebyshev inequility,

$(pe^ rac{x}{p}+qe^ rac{-x}{q})le(p+q) rac{(e^ rac{x}{p}+e^ rac{-x}{q})}{2}$

$(pe^ rac{x}{p}+qe^ rac{-x}{q})le rac{(e^ rac{x}{p}+e^ rac{-x}{q})}{2}$ .................(i)

now, let us assume that $e^ rac{x^2}{8p^2q^2}ge rac{(e^ rac{x}{p}+e^ rac{-x}{q})}{2}$ to be correct,

by AM-GM inequility,

$rac{(e^ rac{x}{p}+e^ rac{-x}{q})}{2}ge e^{ rac{1}{2}( rac{x}{p}- rac{x}{q})}$

so, we have

$e^ rac{x^2}{8p^2q^2} ge e^{ rac{1}{2}( rac{x}{p}- rac{x}{q})}$

as,e^x is an increasing function,

so, $rac{x^2}{8p^2q^2} ge { rac{1}{2}( rac{x}{p}- rac{x}{q})}$

or, $rac{x^2}{8p^2q^2} ge { rac{q-p}{2pq}$ x

which is always true as a positive number is always greater than a negative number as ,

hence our assumption, $e^ rac{x^2}{8p^2q^2}ge rac{(e^ rac{x}{p}+e^ rac{-x}{q})}{2}$ , is correct......

so from (i) we have,

$(pe^ rac{x}{p}+qe^ rac{-x}{q})le e^ rac{x^2}{8p^2q^2}$

now, is their any mistake in it...............if yes, then pls correct it.........

Rajat Sen

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24 Jan 2009 11:15:50 IST

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The function is not exactly symmetric with respect to p,q.

So, your assumption is not right, there is l;oss of generality.

Dipanjan

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24 Jan 2009 13:15:31 IST

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Yes Rajat is right that the solution is wrong.

But now it only remains to prove the inequality using the assumption that p<q.......

Ady JEE-09 AIR 294

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9 Feb 2009 17:03:26 IST

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actually there is no loss of generality, simply replace x by -x as eqn one is true eqn 2 is equivalently true, add the new equation to the previous, you get an equation which is symmetric with respect to p and q.hence without loss of generality you can assume p<q or p>q for the component equations

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