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?

To minimise
\begin{align}
y & = 2^{\cos^2 x} + 2^{\sin^2 x} \\
\frac{dy}{dx} & = \frac{d(2^{\cos^2 x})}{dx} + \frac{d(2^{\sin^2 x})}{dx}
\end{align}
we want $\frac{dy}{dx} = 0$. For the first summand we have
\begin{align}
f(x) & = 2^{\cos^2 x} \\
\ln f(x) & = \cos^2 x\ln 2 \\
\frac{d(\ln f(x))}{dx} & = \frac{d(\cos^2 x\ln 2)}{dx} \\
\frac{f'(x)}{f(x)} & = -2\ln 2\cos x\sin x \\
\frac{d(f(x))}{dx} & = -2^{\cos^2 x}2\ln 2\cos x\sin x
\end{align}
Analogous for the second summand we have $\frac{d(2^{\sin^2 x})}{dx} = 2^{\sin^2 x}2\ln 2\sin x\cos x$ (check). Now, we simply put these back in our original equation
\begin{align}
\frac{dy}{dx} & = -2^{\cos^2 x}2\ln 2\cos x\sin x + 2^{\sin^2 x}2\ln 2\sin x\cos x \\
2^{\cos^2 x}2\ln 2\cos x\sin x & = 2^{\sin^2 x}2\ln 2\sin x\cos x
\end{align}
Here, verify that $x = k\frac{\pi}{2}, k = 1, 2, 3, \dots$ are maxima. Accounting for those, we can now safely cross out
\begin{align}
2^{\cos^2 x} & = 2^{\sin^2 x} \\
\cos^2 x & = \sin^2 x \\
\cos^2 x & = 1 - \cos^2 x \\
\cos^2 x & = \frac{1}{2} \\
\cos x & = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2} \\
x & = \arccos{\pm\frac{\sqrt{2}}{2}} \\
x & = \pm\frac{\pi}{4}
\end{align}
