# Maximium of area/perimeter^2 of a function

Let $a,b$ be two reals. Does there exists a constant $C$ such that for all functions $f:[a,b]\to\mathbb{R}$, continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)=0$, $$\int_a^bf(t)dt\le C\left(\int_a^b\sqrt{1+f'(t)^2}dt\right)^2$$ If there exists such constant, does it depends on $a$ and $b$?

$\int_a^bf(t)dt$ is obviously the area under the graph of $f$ and $\int_a^b\sqrt{1+f'(t)^2}dt$ is meant to be the perimeter of $f$.

I think that $C=\frac{1}{2\pi}$.

-

If you assume $f(t)>0$ for $t\in (a,b)$, and then reflect the graph over the $x$-axis, the graph and its reflected image are the boundary of the region under the graph and its reflected image. Now you can use the isoperimetric inequality to conclude $4\pi A \le L^2$, where $A=2\int_a^b f(t) \, dt$ is the enclosed area and $L=2\int_a^b \sqrt{1+f'(t)^2} \, dt$ is the length of the perimeter. This gives exactly your inequality with the conjectured value of $C=\frac1{2\pi}$. Equality holds if and only if the graph is a semicircle.