 Solution

Rolle's theorem states that a continuous function $g:[u,v]\to \mathbb{R}$ with $u<v$, differentiable on $(u,v)$, and satisfying $g(u)=g(v)$ has $g^{\prime }(c)=0$ for some $c\in (u,v)$.

Fix $x>0$ and set $r=(\sqrt{1+x}-1)/x$. The functionis continuous and differentiable as required, and $g(0)=g(x)=0$. By Rolle's theorem, some $c\in (0,x)$ satisfiesSince $c>0$, this gives $r<1/2$. Multiplying by $x>0$ and adding one proves the strict square root bound