# Logarithmic form of Young's inequality

If $\frac {1}{p}+\frac {1}{q}=1$, Prove: $\log (\frac {x^p}{p}+\frac {y^q}{q})$ bigger or equal to $\frac {1}{p} \log (x^p) +\frac {1}{q} \log (y^q)$ Insightful proof would be appreciated! How can this be prove without Young's inequality?

The Problem is the number 2.22 in the book called "A problem book in real analysis" by Asuman Güven Aksoy, Mohamed A. Khamsi.

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What have you tried? –  Alex Becker Dec 28 '11 at 22:15
@smanoos did most of the TeXifying. I just fixed a couple of parentheses. Re question: Do we know anything about $x$ and $y$? Seems a bit strange that they only appear raised to powers $p$ and $q$ respectively. That is kind pointless, unless something extra is known about them? –  Jyrki Lahtonen Dec 28 '11 at 22:26
Victor, please check that the two numbers are correctly written here. The latter simplifies to $\log (xy)$ and the former to $\log\left(\frac{x^py^q}{pq}\right)$. Here we can forget the logarithms, so in the natural test case $p=q=2$ we get ... –  Jyrki Lahtonen Dec 28 '11 at 22:36
Exponentiation of both sides shows this statement is the generalized AM-GM inequality. One of the clearest proofs uses Jensen's Inequality, as shown in the reply by @VSJ. –  whuber Dec 29 '11 at 16:02

Since $\log$ is a concave function (and $\frac{1}{p} + \frac{1}{q} = 1)$, using Jensen's Inequality we get
$$\log (\frac {x^p}{p}+\frac {y^q}{q}) \geq \frac{1}{p}\log(x^p) + \frac{1}{q}\log(y^q) = \log(xy)$$ This inequality is better known as Young's Inequality.