﻿ Prove the inequality 1 \leq \sqrt{\sin x}+\sqrt{\cos x} \leq 2^{3/4}

#### Question

Prove the inequality 1 \leq \sqrt{\sin x}+\sqrt{\cos x} \leq 2^{3/4}

Collected in the board: Trigonometry

Steven Zheng posted 1 year ago

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)
2\sin x\cos x=(\sin x+\cos x)^2-1

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.

Steven Zheng posted 1 year ago

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