Prove that $\left(\frac{a}{p}\right)^p\left(\frac{b}{q}\right)^q\leq\left(\frac{a+b}{p+q}\right)^{p+q}$

Question

Suppose $p,q> 0$ and $a,b\geq 0$, prove the following inequality

$\left(\frac{a}{p}\right)^p\left(\frac{b}{q}\right)^q\leq\left(\frac{a+b}{p+q}\right)^{p+q}$

I tried taking log on both sides but this does not make things easier. Is there a more neat way to tackle this problem?

I'd say rewrite it $(p+q)^{p+q}a^pb^q\le (a+b)^{p+q}p^pq^q$, start with $a=b=p=q=1$ and take partial derivatives on each side (maybe a bit tedious, but tried and true.) — Tim Ratigan, Dec 05 '13 at 07:01
@TimRatigan So consider both sides as functions of four variables? — Kato yu, Dec 05 '13 at 07:03
@TimRatigan Well..I think this would be painful, but maybe doable — Kato yu, Dec 05 '13 at 07:06

score 6 · Accepted Answer · answered Dec 05 '13 at 07:17

6

Taking logs as you tried, we have to show $$p \log \frac{a}p + q\log \frac{b}q \le (p+q) \log \frac{a+b}{p+q}$$

As $f(x) = \log x$ is concave, using Jensen's inequality we have $$\frac{p}{p+q}f\left(\frac{a}p\right) + \frac{q}{p+q}f\left(\frac{b}q\right)\le \log \left(\frac{p}{p+q}\frac{a}{p}+\frac{q}{p+q}\frac{b}{q}\right) = \log\frac{a+b}{p+q}$$ which is the same as above.

answered Dec 05 '13 at 07:17

Macavity

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It is exactly what we call [log sum inequality] (en.wikipedia.org/wiki/Log_sum_inequality) @Katoyu – Shuchang Dec 05 '13 at 07:25

Prove that $\left(\frac{a}{p}\right)^p\left(\frac{b}{q}\right)^q\leq\left(\frac{a+b}{p+q}\right)^{p+q}$

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