Prove $\sqrt[n]{n}\leq 1+\frac{2}{\sqrt{n}}$ I am given this statement and I need to prove it. the statement is for all $n\ge 1$: $$\sqrt[n]{n}\leq 1+\frac{2}{\sqrt{n}}$$
I am trying to prove with induction. But I am stuck for step n=k+1, how can I then decompose the step:  $\sqrt[n+1]{n+1}$? 
Thanks for help in advance
 A: Induction does not seem to suggest itself for this kind of task. Try to find out something about $$\left(1+\frac2{\sqrt n}\right)^n\stackrel?\ge n$$ instead (using binomials).
EDIT:
To summarize below comments into this solution, there are (at least) two methods to show $\left(1+\frac2{\sqrt n}\right)^n\ge n$ for all $n\ge 1$.
One is to use binomial expansion 
$$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}.$$
here with $a=1$, $b=\frac2{\sqrt n}$. If $n\ge2$ (noting that of course the remaining case $n=1$ can be checked directly) and we drop the nonnegative summands belonging to all $k>2$, we find 
$$\begin{align}\left(1+\frac2{\sqrt n}\right)^n&\ge{n\choose 0}+{n\choose 1}\frac2{\sqrt n}+{n\choose2}\frac 4n\\&=1+2\sqrt n+2(n-1)\\&=n+(n+2\sqrt n-1)\\&\ge n.\ _\square\end{align}$$
Another way is to make use of the famous and very important Bernoulli inequallity
$$ (1+x)^r\ge 1+rx\quad\text{for all }x\ge-1, r\ge 0.$$
If we plug in $x=\frac2{\sqrt n}$ (of course) and $r=\frac n2$ (a bit surprisingly), we obtain
$$\left(1+\frac2{\sqrt n}\right)^{\frac n2} \ge 1+\sqrt n$$
and by squaring this
$$\left(1+\frac2{\sqrt n}\right)^{n} \ge (1+\sqrt n)^2=1+2\sqrt n + n> n.\ _\square$$
A: I think Hagen's answer is better than this one, but this one illustrates a useful general principle: take logarithms to simplify exponentials (and products). A second useful principle is to use power series to get inequalities for complicated functions in terms of simple polynomials.
In this case, since $\ln$ is an increasing function, the inequality $\sqrt[n]n \le 1 + 2/\sqrt n$ is equivalent to $\ln(\sqrt[n]n) \le \ln(1 + 2/\sqrt n)$, or $\frac1n \ln n \le \ln(1 + 2/\sqrt n)$. Now we use the inequality $\ln(1+x) \ge x - x^2/2$, valid for all $0\le x\le1$; one way to see this is by noting that the power series $\ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + \cdots$ is an alternating series of decreasing terms in that range. Thus it suffices to show that
$$
\frac{\ln n}n \le \frac2{\sqrt n} - \frac12\bigg( \frac2{\sqrt n} \bigg)^2,
$$
or equivalently that
$$
\ln n + 2 \le 2\sqrt n;
$$
this is much easier to check than the original problem.
