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?

Let $y=2^{\sin^2x}+2^{\cos^2x}=2^{\sin^2x}+2^{1-\sin^2x}$
$$(2^{\sin^2x})^2-y\cdot2^{\sin^2x}+2=0$$ which is a Quadratic Equation in $2^{\sin^2x}$
So, the discriminant must be $\ge0$
$$(y)^2\ge4\cdot2\implies y^2\ge8$$
As $y>0,y\ge2\sqrt2$
The equality occurs if $$2^{\sin^2x}=\dfrac{2\sqrt2}2=\sqrt2=2^{1/2}$$
i.e., if $\sin^2x=\dfrac12\iff\cos2x=0$
