Question

Prove that $2^{\sin x}+2^{\cos x}\le 2^{1-\frac{1}{\sqrt{2}}}$ for all real $x$.

Answer 1

$2^{\sin x}+2^{\cos x}\le 2{-}^{1-\frac{1}{\sqrt{2}}}$ $x\in R$

by $AM\ge GM$ in equality,

$\frac{2^{\sin x}+2^{\cos x}}{2}\ge 2^{\frac{1}{2}(\sin x+\cos x)}$

$2^{\sin x}+2^{\cos x}\ge {2.2}^{\frac{1}{2}(\sin x+\cos x)}$ .....$\left(1\right)$

since $\sin x+\cos x$

$=\sqrt{2}(\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}})$

$=\sqrt{2}\sin (x+\frac{\pi }{4})$ ........$\left(2\right)$

From eq $\left(1\right)$ & eq $\left(2\right)$

R.H.S$\Rightarrow {2.2}^{\frac{1}{2}(\sin x+\cos x)}$

$=2^{1+\frac{1}{2}\left(\sqrt{2}\sin (x+\frac{\pi }{4})\right)}$

$=2^{1+\frac{1}{\sqrt{2}}\sin (x+\frac{\pi }{4})}$

$\ge 2^{1-\frac{1}{\sqrt{2}}}$

as $-1\le \sin (x+\frac{\pi }{4})\le 1$

$2^{\sin x}+2^{\cos x}\ge 2^{1+\frac{1}{\sqrt{2}}\sin (x+\frac{\pi }{4})}\ge 2^{1-\frac{1}{\sqrt{2}}}$.