Find the minimum value. 
Find  the minimum value of $4^x + 4^{1-x}$  , $x\in\mathbb{R}$.

In this I used the property that $a + \frac{1}{a}\geq 2$.
So I begin with 
$$ 4^x + \left(\frac{1}{4}\right)^x + 3\left(\frac{1}{4}\right)^x \geq 2 + 3\left(\frac{1}{4}\right)^x$$
So I think the minimum value be between 2 to 3.
But the answer is 4.
Thanks in advance.

$$4^{x} + 4^{1-x} = (\sqrt{4^x} - \frac{2}{\sqrt{4^x}})^2 + 4 \ge 4$$
with equality occuring when $\sqrt{4^x} = \frac{2}{\sqrt{4^x}}$ and so $x = \frac{1}{2}$.
