Generalization of $\int_0^\alpha \sqrt{1+\cos^2\theta}\,d\theta>\sqrt{\alpha^2+\sin^2\alpha}$

The inequality naturally only makes sense for continuously differentiable $f$. We will use that if $F$ and $G$ are continuously differentiable, and $F(0)=G(0)$ and $F'(y) \geq G'(y)$ for all $y \in [0,z]$ then $F(y) \geq G(y)$ on that interval.

Equality holds for $a=0$. Differentiating both sides w.r.t $a$ gives $$\sqrt{1+(f'(a))^2} \geq \frac{a+f(a)f'(a)}{\sqrt{a^2+f(a)^2}}$$ Which after squaring both sides and rearranging amounts to $$a^2f'(a)^2 + f(a)^2 \geq 2a f(a) f'(a)$$ or $$ (f(a)-a f'(a))^2=0.$$ Insisting that this holds for all $a$ is easily seen to imply linearity of $f$.

My "geometrically" solution.

https://i.stack.imgur.com/wYgjx.png