Minimum value of $2^{\sin^2x}+2^{\cos^2x}$ The question is what is the minimum value of
$$2^{\sin^2x}+2^{\cos^2x}$$
I think if I put $x=\frac\pi4$ then I get a minimum of $2\sqrt2$. But how do I prove this?

We have that
$$\min_{x\in\mathbb{R}}\left\{2^{\sin^{2}x}+2^{\cos^{2}x}\right\}=
\min_{t\in[0,1]}\left\{2^{t}+2^{1-t}\right\}=\min_{r\in[1,2]}\left\{r+\frac{2}{r}\right\}=2\sqrt{2}$$
where in the last step we used the fact that for $r>0$, 
$$r+\frac{2}{r}\geq 2\left(r\cdot\frac{2}{r}\right)^{1/2}=2\sqrt{2}$$ and the equality holds if $r=\sqrt{2}\in[1,2]$.
