<s>[INST] If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$. Need help with proof. If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$.
Working:
$\frac{1}{2}(1-a)\leq 0<\frac{1}{2}(b-a)$ and $(1-\sqrt{a})\leq 0<\sqrt{b}-\sqrt{a}$
Trying to show $\frac{1}{2}(b-a)-(\sqrt{b}-\sqrt{a})$ is positive but don't know what to do next.
 [/INST] One may observe that, for $b>a\ge1$,
$$
\begin{align}
(b-a)-2(\sqrt{b}-\sqrt{a})&=(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})-2(\sqrt{b}-\sqrt{a})
\\\\&=(\sqrt{b}-\sqrt{a})\left[\sqrt{b}+\sqrt{a}-2 \right]
\\\\&=(\sqrt{b}-\sqrt{a})\left[(\sqrt{b}-1)+(\sqrt{a}-1) \right]
\\\\&>0
\end{align}
$$ by using that $x \mapsto \sqrt{x}$ is strictly increasing over $[1,\infty)$.
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