Opial type inequalities Let $x(t)\in C^1[0,h]$ be such that $x(0)=x(h)=0$ and $x(t)$ in (0,h) ,then the following inequality
$ \int^h_0 |x(t)x^{'}(t)|dt \leq \frac{h}{4}\int^h_0(x^{'}(t))^2dt$
my question: I would like proof of this inequality
 A: For $t\in [0,h/2]$ write $|x(t)|=|\int_0^t x'(s)ds|\leqslant \int_0^t |x'(s)|ds$, thus $$\int_0^{h/2}|x(t)x'(t)|dt\leqslant \int\int_{0\leqslant s\leqslant t\leqslant h/2}|x'(s)|\cdot |x'(t)|dt=\frac12 \int_0^{h/2}\int_0^{h/2}|x'(s)|\cdot |x'(t)|dsdt\leqslant \\\frac14\int_0^{h/2}\int_{0}^{h/2}|x'(s)|^2+|x'(t)|^2 dsdt=\frac{h}4\int_0^{h/2}|x'(s)|^2 ds,$$
analogously
$$\int_{h/2}^{h}|x(t)x'(t)|dt\leqslant \frac{h}4\int_{h/2}^{h}|x'(s)|^2 ds,$$
sum up.
A: Define $y(t)=x(ht/2)$ so that $y(0)=y(2)=0$. It’s easy to see that the inequality for $x$ holds if and only if 
$$
\int_0^2 |2y(t)\dot y(t)|\,dt\le \int_0^2 |\dot y(t)|^2\, dt.
$$
The left side is exactly the variation of $y^2$ over the interval [0,2]. We will check the inequality in the case where $y$ has finitely many intervals of monotonicity. 
Suppose the coordinates in the plane of the endpoints of these intervals on the graph are $(a_0,b_0),\ldots (a_{2n},b_{2n})$ where even points are local maxima and odd points are local minima.
If the function is rearranged with the increasing intervals first, and the decreasing intervals later, then the left side becomes larger (as $y(t)$ is larger), but the right side stays the same. Hence it suffices to prove the inequality for unimodal functions. 
Suppose the function increases from $(0,0)$ to $(a,b)$ and then decreases to $(2,0)$. The left side is $2b^2$. By Jensen’s inequality, the integral of $y’^2$ given that $\int_0^a y’=b$ is minimized when $y$ is linear, so that the contribution to the right side of the inequality from the interval $[0,a]$ is at least $b^2/a$. The contribution from $[a,2]$ is similarly at least $b^2/(2-a)$. The sum is minimized when $a=1$. Hence the right side of the inequality is at least $2b^2$ and so the inequality is proved in the unimodal case, and hence in the case with finitely many intervals of monotonicity. The general case follows by approximation. 
Note: there is likely a much cleaner proof of this fact out there somewhere. 
