Question
Prove the inequality 1 \leq \sqrt{\sin x}+\sqrt{\cos x} \leq 2^{3/4}
Answer
1 \leq \sqrt{\sin x}+\sqrt{\cos x} \leq 2^{3/4}
(1)
1 \leq \sin x+\cos x+2\sqrt{\sin x\cos x} \leq 2^{\tiny3/2}
(2)
Let t = \sin x+\cos x
\sin x\cos x = \dfrac{1}{2} t^2-\dfrac{1}{2}
The inequality (2) is converted to
1 \leq t+2\sqrt{ \dfrac{1}{2} t^2-\dfrac{1}{2}} \leq 2^{\tiny3/2}
(3)
On the other hand,
t = \sin x+\cos x
=\sqrt{2} (\dfrac{\sqrt{2} }{2} \sin x+ \dfrac{\sqrt{2} }{2} \cos x)
= \sqrt{2} \sin (x+\dfrac{\pi}{4} )
And \dfrac{1}{2} t^2-\dfrac{1}{2}\geq 0
t\geq 1 or t\leq -1
From (1), we know
\sin x \geq 0 and \cos x \geq 0 , so t \geq 0
Therefore, the domain of t in function (2) is ,
1 \leq t \leq \sqrt{2}
As function f(t) = t+2\sqrt{ \dfrac{1}{2} t^2-\dfrac{1}{2}} is a monotonic increasing function in the domain of [1,\sqrt{2} ]
Therefore it has
maximum value f(\sqrt{2}) = 2^{\tiny3/2}
and minimum value f(1) = 1
hence, the inequality (3) is proved.