Let $w >0$ and let $f: [−w, w]→R$ be a continuous function that is piecewise continuously differentiable on (−w, w). Assume that $f(x)≥0$ for all $x∈[−w, w]$ and $f(−w) =f(w) = 0$. Let a be the area under the graph of $f$ bounded above the x-axis and let $l$ be the length of the graph of $f$.
Show that $\frac{\pi w^2}{2}+a \leq wl $:
We know that $a = \int_{-w}^{w} f(x)\ dx = \int_{-w}^{w} -xf'(x)\ dx $, using integration by parts and $\frac{\pi w^2}{2} = \int_{-w}^{w} \sqrt{w^2-x^2}\ dx$ using u-substitution of $x=wsin(u),\ l =\int_{-w}^{w} \sqrt{1+f(x)^2}\ dx $
Cauchy-Schwarz: $(\int_{-w}^w h g)^2\leq \int_{-w}^w |h|^2\int _{-w}^w|g|^2 \\$. I have tired using $h= 1+f'(x) $ and $ g = -x +\sqrt{w^2-x^2}$
However using these substitutions in Cauchy-Schwarz (given above) I get:
$ \int_{-w}^{w} \sqrt{w^2-x^2}\ dx +\int_{-w}^{w} -xf'(x)\ dx \leq -\int_{-w}^{w} f'(x) \sqrt{w^2-x^2}\ dx+\sqrt{2a^3\int_{-w}^{w} 1+f'(x)^2\ dx}$ . This is same as
$ \frac{\pi w^2}{2} +a \leq -\int_{-w}^{w} f'(x) \sqrt{w^2-x^2}\ dx+\sqrt{2a^3\int_{-w}^{w} 1+f'(x)^2\ dx}$ which is not quite equivalent to the question.
Moreover, using the Cauchy-Schwarz again: $\sqrt{2a^3\int_{-w}^{w} 1+f'(x)^2\ dx}≥a\int_{-w}^{w} \sqrt{1+f'(x)^2}=al $.
Any advice would be highly appreciated, thank you!