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?
 A: 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 $\sin^2x=\dfrac12\iff\cos2x=0$
A: 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 $x = \arcsin \frac{1}{\sqrt{2}} = 45^\circ$
A: 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 $r=\sqrt{2}\in[1,2]$.
A: By the AM-GM inequality
$$ 2^{\sin^2(x)}+2^{\cos^2(x)} \geq 2\sqrt{2^{\sin^2(x)}\cdot 2^{\cos^2(x)}} =2\sqrt{2}$$
and equality is achieved only when $2^{\sin^2(x)}=2^{\cos^2(x)}$, i.e. only when $\sin^2(x)=\cos^2(x)$.
A: As always with minima, take a derivative and set it equal to zero, then solve that equation.
A: 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}
