Let $a,b>0$, $f(0)=0$, $f(a)=b$, $f(x)\geq 0$ and $f''(x)\geq 0$ for all $x$. Show that $$2\int_0^a f(x)\sqrt{1+(f')^2(x)}\,dx\leq b\sqrt{a^2+b^2}.$$
It is easy to see its geometry, the area of $f$ revolved along the $x$-axis. But how to estimate? $f$ is easy by convexity, but $f'$?
Just an idea only as I have no time to work out the details.
Define $F:[0,\infty)\rightarrow\mathbb{R}$ by $$ F(t)=f(t)\sqrt{t^{2}+f^{2}(t)}-2\int_{0}^{t}f(x)\sqrt{1+f'^{2}(x)}dx. $$ Clearly $F$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. Moreover, $F(0)=0$. Investigate the derivative $F'(t)$. If you can prove that $F'(t)\geq0$, then you can conclude that $F$ is increasing and hence $F(t)\geq F(0)=0$. In particular, $F(a)\geq0$.