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?

Solution 1:

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 $r=\sqrt{2}\in[1,2]$.

Solution 2:

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 $\sin^2(x)=\cos^2(x)$.

Solution 3:

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 $\sin^2x=\dfrac12\iff\cos2x=0$

Solution 4:

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 $x = \arcsin \frac{1}{\sqrt{2}} = 45^\circ$