TITLE: Prove or disprove the inequality $\sqrt{1+\sin b}-\sqrt{1+\sin a}\leq\frac{b-a}{2}$
QUESTION [12 upvotes]: I'm asked to prove or to disprove the inequality $$\sqrt{1+\sin b}-\sqrt{1+\sin a}\leq\frac{b-a}{2}$$
for all $0<a<b<\pi/2$.
I believe that it is true. 
Proof: Let $f(x):=\sqrt{1+\sin x}$. The function $f(x)$ is continuous on the interval $[a,b]\subseteq[0,\pi/2]$ and is differentiable on $(a,b)$. Thus, by the mean value theorem there exists a point $c\in (a,b)$ such that
$$f'(c)=\frac{\cos c}{2\sqrt{1+\sin c}}=\frac{f(b)-f(a)}{b-a}$$
and because $0<c<\pi/2$,
$$\frac{\cos c}{2\sqrt{1+\sin c}}<\frac{\cos 0}{2\sqrt{1+\sin 0}}=\frac{1}{2}$$
And thus:
$$f'(c)=\frac{f(b)-f(a)}{b-a}<\frac{1}{2} \Rightarrow \sqrt{1+\sin b}-\sqrt{1+\sin a}\leq\frac{b-a}{2}$$
Is it correct?

REPLY [3 votes]: Your approach works. If you like an alternative way,
$$ f(x)=\sqrt{1+\sin x} = \sqrt{2}\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)=\sqrt{2}\sin\left(\frac{\pi}{4}+\frac{x}{2}\right) $$
gives that $f(x)$ is an increasing and concave function over $\left[0,\frac{\pi}{2}\right]$ with Lipschitz constant $\frac{1}{2}$, hence:
$$ \sqrt{1+\sin b}-\sqrt{1+\sin a}\leq\frac{b-a}{2} $$
as wanted.