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$p$ and $q$ are positive interger and $\frac{1}{p}+\frac{1}{q}=1$

For $0 \le u$ and $0\le v$
prove that $uv \le \frac{u^p}{p}+\frac{v^q}{q}$.

Put $f(x)=x^{p-1}, f^{-1}=y^{q-1}$ then
$$uv \le \int_{0}^{u^p} f(x)dx+ \int_{0}^{v^q} f^{-1}(y)dy$$

Why this inequality holds?

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This is Young's Inequality. A proof (which I haven't checked), can be found here. – David Mitra Dec 15 '12 at 13:46
Yes, I just want to know why that integral calculation is approved in here. I'll check your link. Thanks! – landolf Dec 15 '12 at 13:51
Ah... The informal proof is to interpret the inequality in terms of areas; per the diagrams in the links above. – David Mitra Dec 15 '12 at 13:58
Do you have the correct upper limits in the integrals? – David Mitra Dec 15 '12 at 14:03
A formal proof of the generalized form of the integral inequality can be found here. Not that $uv\le {u^p\over p}+{v^q\over q}$ is a consequence of the integral inequality, but can also be proved more simply as in proof 1 in the proofwiki link above. – David Mitra Dec 15 '12 at 14:22
up vote 0 down vote accepted

A simple calulation shows that $$\int_{0}^{u^p} f(x)dx+ \int_{0}^{v^q} f^{-1}(y)dy=\int_{0}^{u^p} x^{p-1}dx+ \int_{0}^{v^q} y^{q-1}dy=\frac{x^p}{p}|_{0}^{u^p}+\frac{y^q}{q}|_{0}^{v^q}=\\ \frac{(u^p)^p}{p}+\frac{(v^q)^q}{q} $$ By the Young Inequality, $$\frac{(u^p)^p}{p}+\frac{(v^q)^q}{q}\ge u^pv^q $$ Perhaps you meant $$\int_{0}^{u} f(x)dx+ \int_{0}^{v} f^{-1}(y)dy\ge uv $$ ?

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