Proving that $\frac{u^p}{p}+\frac{v^q}{q}\ge uv$ under the condition $\frac{1}{p}+\frac{1}{q}=1$ The following is a problem (6.10) from Rudin's principles of Mathematical analysis.

Let $p$ and $q$ be positive real numbers such that $$\frac{1}{p}+\frac{1}{q}=1.$$ Prove that if $u\ge 0$ and $v\ge 0$, then $$uv\le \frac{u^p}{p}+\frac{v^q}{q}.$$

I can prove this by using Weighted Arithmetic mean Geometric mean inequality and also by using Jensen's inequality on natural logarithm (this is usually used to the prove generalized AM-GM). 
I would like to see alternate elementary methods (preferably avoiding multivariate calculus methods)  to solve this (I think Rudin hasn't introduced convexity before this; so generalized AM-GM is cheating).
 A: An idea: Assume $p > 1$ and write the relation $ \frac{1}{p} + \frac{1}{q} = 1 $ as follows:
$(p-1)(q-1) = 1 $. Put $x = t^{p-1}$. Hence, $t = x^{q-1}$. the case when $u,v$ are $0$ is a triviality. Hence, assume $u > 0 $ and $v > 0$. Consider in the $xt-$plane the rectangle furnished by the lines $x = v$ and $t = u.$ Draw and picture a show that 
$$ uv \leq \int\limits_0^v t^{p-1} dt + \int\limits_0^u x^{q-1} dx $$
The right hand side is obviously $$\frac{ v^p}{p} + \frac{u^q}{q}$$
A: First of all, Rudin never cheats. Before studying his book, you must have studied basic calculus, so that convexity by means of second derivatives should be already known.
Anyway, and suggested in a comment by Alex H., a standard way to prove this inequality is to show that
$$
\frac{v^q}{q} = \max_{u \geq 0} \left( uv - \frac{u^p}{p} \right),
$$
by using standard calculus. Theoretically, you are computing the Fenchel transform of $u \mapsto u^p/p$, which is a powerful tool of Convex Analysis.
This approach is suggest in the exercises of Michel Willem's book Functional analysis, published by Birkhäuser.
As a comment, we should agree that this is a convexity inequality, as you can read in Willem's book: this means that it is a particular case of a general inequality about convex functions. It is hardly believable that you can prove it without any explicit or implicit reference to the idea of convexity.
A: Let $f(x)=x^{p-1}$ and $g(x)=x^{q-1}$. Then:


*

*$g=f^{-1}$;

*$\displaystyle\int_0^uf(x)\,dx=\frac{u^p}p$;

*$\displaystyle\int_0^vg(x)\,dx=\frac{v^p}p$.


Therefore, you want to prove that$$uv\leqslant\int_0^uf(x)\,dx+\int_0^vf^{-1}(x)\,dx.$$This is clear graphically, since the graph if $f^{-1}$ can be obtained from the graph of $f$ by reflection on the line $y=x$ and since $uv$ is the area of the rectangle whose vertices are $(0,0)$, $(u,0)$, $(u,v)$, and $(0,v)$.
