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?

You know that $\cos^2 x = 1 - \sin^2x$, so you can rewrite:
$$
2^{\sin^2x} + 2^{\cos^2x} = 2^{\sin^2x} + 2^{1-\sin^2x} = 2^{\sin^2x} + \frac{2}{2^{\sin^2x}}
$$
Now, let $2^{\sin^2 x} = y$, then we basically have to maximize $y + \frac{2}{y}$. But then, note that: $y + \frac{2}{y}$ has derivative $1 - \frac{2}{y^2}$, which is $0$ when $y^2 = 2$ or $y = \sqrt{2}$. Hence, $\sin^2 x = \frac{1}{2}$, hence $x = \arcsin \frac{1}{\sqrt{2}} = 45^\circ$
