Trigonometric problem: $2^{\sin{x}} + 2^{\cos{x}} \ge 2^{(1-1/{\sqrt2})}$ Show that:
$$\large2^{\sin{x}} + 2^{\cos{x}} \ge 2^\left({1-\frac{1}{\sqrt{2}}}\right)$$
This looks like an am gm problem to me where we should be using the fact that am is more that or equal to gm but I am having problem solving this equation after this:
$\large\left(2^{\sin{x}} + 2^{\cos{x}}\right)/2 \ge \sqrt{2^{\sin{x}+\cos{x}}}$
Please solve the sum step wise. Thanks

By the am-gm inequality,
\begin{align}
\frac{2^{\sin x}+2^{\cos x}}{2}
&\ge 2^{\frac{1}{2}(\sin x+\cos x)}\\
\end{align}
Multiplying both sides of the inequality by $2$ yields
$$2^{\sin x}+2^{\cos x}\ge2\cdot 2^{\frac{1}{2}(\sin x+\cos x)}$$
Since $\color\red{\sin{x}+\cos{x}=\sqrt{2}\sin(x+\pi/4)}$, we have
\begin{align}
RHS
&=2\cdot2^{\frac{1}{2}(\sin{x}+\cos{x})}\\
&=2^{1+\frac{1}{\sqrt{2}}\sin(x+\pi/4)}\\
&\ge2^{1-\frac{1}{\sqrt{2}}}
\end{align}
as $-1\le \sin(x+\pi/4)\le 1.$
Hence
$$\color\red{2^{\sin x}+2^{\cos x}}\ge 2^{1+\frac{1}{\sqrt{2}}\sin(x+\pi/4)}\ge \color\red{2^{1-\frac{1}{\sqrt2}}}$$
