= Solution
{c}
\b[<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 $g'(c)=0$ for some $c\in(u,v)$.
Fix $x>0$ and set $r=(\sqrt{1+x}-1)/x$. The <function>
$$
g(t)=\sqrt{1+t}-1-rt,\qquad0\leq t\leq x,
$$
is <continuous> and <differentiable> as required, and $g(0)=g(x)=0$. By <Rolle's theorem>, some $c\in(0,x)$ satisfies
$$
0=g'(c)=\frac1{2\sqrt{1+c}}-r.
$$
Since $c>0$, this gives $r<1/2$. Multiplying by $x>0$ and adding one proves the strict <square root> bound
$$
\boxed{\sqrt{1+x}<1+x/2\qquad(x>0).}
$$
Back to article page
