How do I prove a trigonometric inequality?

[anemone]

Prove that for all real numbers $x$, we have \(\displaystyle \left(2^{\sin x}+2^{\cos x}\right)^2\ge2^{2-\sqrt{2}}\).

 

[Greg]

Prove that for all real numbers $x$, we have \(\displaystyle \left(2^{\sin x}+2^{\cos x}\right)^2\ge2^{2-\sqrt{2}}\).

Spoiler

AM-GM inequalty (as the LHS is always positive) with square root of LHS:

$$\dfrac{2^{\sin{x}}+2^{\cos{x}}}{2}\ge\sqrt{2^{\sin{x}}\cdot2^{\cos{x}}}=2^{\dfrac{\sqrt2}{2}\sin\left(x+\dfrac{\pi}{4}\right)}$$

$$2^{\sin{x}}+2^{\cos{x}}\ge2^{\dfrac{\sqrt2}{2}\sin\left(x+\dfrac{\pi}{4}\right)+1}$$

$$\min\left(\dfrac{\sqrt2}{2}\sin\left(x+\dfrac{\pi}{4}\right)+1\right)=1-\dfrac{\sqrt2}{2}$$

Square root of RHS:

$$2^{1-\dfrac{\sqrt2}{2}}$$

hence proved (with equality at $x=\dfrac{5\pi}{4}+2k\pi,k\in\mathbb{Z}$).

 

[anemone]

Good job, greg1313!