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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 $f(a)=f(b)=0$, \begin{equation} \int_a^bf(t)dt\le C\left(\int_a^b\sqrt{1+f'(t)^2}dt\right)^2 \end{equation} 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 $C=\frac{1}{2\pi}$.

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up vote 3 down vote accepted

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.

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Well, you have used a bazooka there. I guess there are no simple solutions to my question. –  xD13G0x Oct 16 '12 at 2:09
    
Since the statement is essentially equivalent to the isoperimetric inequality, I don't think there will be any simpler proofs than those that prove the full isoperimetric inequality. I don't think the graph structure and reflection symmetry will make the statement easier to prove. –  Lukas Geyer Oct 16 '12 at 2:18

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