Proving that: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$ Let $a$ and $b$ be positive reals. Show that
$$\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$$

You can use the following inequality:
$$ \sqrt{xy} \le \frac{x+y}{2} \le \sqrt[x+y]{x^x y^y}$$
The first inequality is straightforward, and the second one can be gotten by
$$ \frac{2}{x+y} = \frac{ x \times 1/x + y \times 1/y}{x+y} \ge \sqrt[x+y]{\frac{1}{x^x y^y}}$$
using the weighted $\text{AM} \ge \text{GM}$.
Setting $x = a^{1/n}$, $y = b^{1/n}$ and taking the $n^{th}$ powers gives us that the limit is $\sqrt{ab}$, by the squeeze theorem.
