Prove that $\,\sqrt [n] n < 1 + \sqrt{\frac{2}{n}}\,$ I am having difficulty proving the following inequality:
$$
\sqrt[n]{n} < 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n.
$$
I am trying to use mathematical induction but I am having trouble going from the left side to the right side (in the induction step) and vice versa. 
I also tried to take the power of $n$ on both sides and somehow use the binomial expansion but I am having difficulty with that. Thank you!
 A: We shall show that $\,\Big(1+\sqrt{\frac{2}{n}}\,\Big)^n>n$. 
For $n=1$ it is obvious. 
Assume that $n\ge 2$, then according to the Binomial Formula:
\begin{align}
\left(1+\sqrt{\frac{2}{n}}\right)^n&=\binom{n}{0}+\binom{n}{1}\sqrt{\frac{2}{n}}+\binom{n}{2}\left(\sqrt{\frac{2}{n}}\right)^{\!2}+\cdots+\binom{n}{n}\left(\sqrt{\frac{2}{n}}\right)^{\!n} \\
&\ge 1+n\sqrt{\frac{2}{n}}+\frac{n(n-1)}{2}\cdot\frac{2}{n}=n+\sqrt{2n}>n.
\end{align}
A: We want to show: $ \sqrt[n]{n} < \frac{ n + \sqrt{2n} } { n} $.
Recognizing this in it's AM-GM form, we want to find $n$ terms whose product is $n$, and whose sum is (at most) $ n + \sqrt{2n}$. With those, we can conclude that
$$ \frac{ n + \sqrt{2n} } { n } \geq \frac{\sum x_i } { n} \geq  \sqrt[n]{\prod x_i } = \sqrt[n]{n}.$$
There are various ways of finding such a set, especially when we restrict to large enough $n$.
EG
For $n \geq 3$, take $ x_1 = x_2 = x_3 = \sqrt[3]{n}, x_4 = \ldots = x_n = 1$.
Verify that $ n + \sqrt{2n} > 3 \sqrt[3]{n} + (n-3)$.
For $n =1, 2$, we can manually check that inequality holds.
