, 2016

We use proof by induction.
For $n = 2$ we have
$$\frac{1}{2} + \sqrt{\frac{1}{2}} < \frac{4}{3} \Leftrightarrow \sqrt{\frac{1}{2}} < \frac{5}{6} \Leftrightarrow \frac{1}{2} < \frac{25}{36},$$
which is true. For $n = 2$ the inequality holds.

Now assume that the inequality holds for $n$ and let us prove it for $n + 1$. We wish to show that
$$\frac{1}{2} + \sqrt{\frac{1}{2}} + \sqrt[3]{\frac{2}{3}} + \dots + \sqrt[n]{\frac{n-1}{n}} + \sqrt[n+1]{\frac{n}{n+1}} < \frac{(n+1)^2}{n+2}. \quad (1)$$
By the induction hypothesis we can estimate
$$\frac{1}{2} + \sqrt{\frac{1}{2}} + \sqrt[3]{\frac{2}{3}} + \dots + \sqrt[n]{\frac{n-1}{n}} + \sqrt[n+1]{\frac{n}{n+1}} < \frac{n^2}{n+1} + \sqrt[n+1]{\frac{n}{n+1}}.$$
It therefore suffices to show that
$$\begin{aligned} \frac{n^2}{n+1} + \sqrt[n+1]{\frac{n}{n+1}} &\le \frac{(n+1)^2}{n+2} \\ \Leftrightarrow \sqrt[n+1]{\frac{n}{n+1}} &\le \frac{(n+1)^2}{n+2} - \frac{n^2}{n+1} = \frac{n^2 + 3n + 1}{(n+2)(n+1)} = 1 - \frac{1}{(n+2)(n+1)}. \end{aligned}$$
Using the Arithmetic-Geometric Mean Inequality we get
$$\sqrt[n+1]{\frac{n}{n+1}} = \sqrt[n+1]{\frac{n}{n+1} \cdot \underbrace{1 \cdot 1 \cdots 1}_{n}} \le \frac{\underbrace{\frac{n}{n+1} + 1 + \cdots + 1}_{n}}{n+1} = \frac{n^2 + 2n}{(n+1)^2} = 1 - \frac{1}{(n+1)^2}.$$
From here it follows that
$$\sqrt[n+1]{\frac{n}{n+1}} \le 1 - \frac{1}{(n+1)^2} < 1 - \frac{1}{(n+2)(n+1)},$$
which proves the inequality (1) and completes the induction step.

Remark: The inequality
$$\sqrt[n+1]{\frac{n}{n+1}} \le 1 - \frac{1}{(n+1)^2}$$
also follows directly from the Bernoulli inequality which states that
$$(1+x)^r \le 1+rx \text{ for all } x > -1 \text{ and } 0 \le r \le 1.$$