Trigonometric problem: $2^{\sin{x}} + 2^{\cos{x}} \ge 2^{(1-1/{\sqrt2})}$

Question

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

Answer 1

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 $$\color\red{2^{\sin x}+2^{\cos x}}\ge 2^{1+\frac{1}{\sqrt{2}}\sin(x+\pi/4)}\ge \color\red{2^{1-\frac{1}{\sqrt2}}}$$

Answer 2

Deriving to find the extrema, you get $$\ln2\cos x\ 2^{\sin x}-\ln2\sin x\ 2^{\cos x}=0,$$ which has the obvious solutions $\sin x=\cos x$, i.e. $x=\frac\pi4+k\pi$.

The corresponding function values are $2\ 2^{\pm1/\sqrt2}$, of which the smallest is $2^{(1-1/\sqrt2)}$. The bound is tight.

(Remains to prove that these are the only roots.)