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If $\frac1{x}+\frac1{y}=1$, for all $ x>1 $ Then $mn \le \dfrac{ym^x+xn^y}{xy}$ with equality $m^x =n^y$, where $m,n \ge 0$.

EDIT: Internet search results say it Hölder's Inequality or a particular form of Hölder's inequality.

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is it Young's inequality? –  newbie Sep 8 '11 at 10:52
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up vote 5 down vote accepted

Young's (thanks @newbie and @AlexanderThumm) inequality states that for non-negative $m,n$,

$$ m\cdot n \leq \frac{1}{x} m^x + \frac{1}{y} n^y $$

where $x, y \geq 1$ and $x^{-1} + y^{-1} = 1$. (This is the inequality you wrote down.)

First notice that in the case where $x= y = 2$, this is nothing by the arithmetic-mean-geometric-mean inequality.

The inequality itself can be seen to be a consequence of Jensen's inequality, as it is equivalent to the statement that (using that $\log$ is a increasing function) (assuming $m\neq 0 \neq n$; if either of $m,n$ is zero, the inequality is trivially true)

$$ \log m + \log n \leq \log \left( \frac{1}{x} m^x + \frac{1}{y} n^y \right) $$

Defining $M = m^x$ and $N = n^y$, this is equivalent to

$$ \frac{1}{x}\log M + \frac{1}{y}\log N \leq \log \left(\frac{1}{x}M + \frac{1}{y} N \right) $$

and this follows from the fact that $\log$ is a concave function.

Using the strict concavity of $\log$ (which you can check by taking its second derivative), you see that the LHS and the RHS equal only when $M = N$, recovering the claim.

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Interesting! I think this answers the problem. Thanks. –  gaurav Sep 8 '11 at 14:01
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This is actually called Young's inequality:

Since $x \mapsto e^x$ is strictly convex ($\frac{d^2}{dx^2}e^x = e^x > 0$) we have

$$mn = e^{\ln m + \ln n} \leq \frac{1}{x}e^{x\ln m} + \frac{1}{y}e^{y \ln n} = \frac{1}{x}m^x + \frac{1}{y}n^y$$

with equality iff $m^x = n^y$.

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Hint:

Fix $n$, and minimize the function $$f(m)=\frac{m^x}{x} + \frac{n^y}{y}-mn.$$
Don't forget to use the condition $\frac{1}{x}+\frac{1}{y}=1.$

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