Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality: 
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...
 A: Inequality Requested
$$
\begin{align}
\left(1+\frac1{\sqrt{n}}\right)^{2n}
&\ge\left(1+\sqrt{n}\right)^2\tag1\\
&\ge n\tag2
\end{align}
$$
Explanation:
$(1)$: Bernoulli's Inequality
$(2)$: $1+\sqrt{n}\ge\sqrt{n}$

Slightly Stronger Inequality
$$
\begin{align}
\left(1+\sqrt{\frac2n}\right)^n
&\ge1+n\sqrt{\frac2n}+\frac{n(n-1)}2\frac2n\tag3\\
&\ge n\tag4
\end{align}
$$
Explanation:
$(3)$: Binomial Theorem
$(4)$: $n+\sqrt{2n}\ge n$
A: Beware: overkill incoming. We may use the AM-GM inequality for producing tight bounds for $\sqrt[n]{n}$. In particular, by noticing that
$$ n = 1\cdot \frac{2}{1}\cdot\frac{3}{2}\cdot\ldots\cdot\frac{n}{n-1} $$
we have
$$ \sqrt[n]{n}=\text{GM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right)<\text{AM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right) $$
and 
$$ \sqrt[n]{n} < 1+ \frac{H_{n-1}}{n},\qquad \left(1+\frac{1}{\sqrt{n}}\right)^2=1+\frac{2\sqrt{n}+1}{n}, $$
so it is enough to show that $H_{n-1}\leq 2\sqrt{n}+1$, at least from some point on. On the other hand the Cauchy-Schwarz inequality ensures
$$ H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}\leq\sqrt{\sum_{k=1}^{n-1}1\sum_{k=1}^{n-1}\frac{1}{k^2}} < \sqrt{n\zeta(2)} $$
and we are done.
A: This is the same as $n\le (1+1/\sqrt n)^{2n}.$ Expanding  using the binomial formula, we want
$$n\le \sum_{k=0}^{2n}\binom{2n}{k}1^{2n-k}(1/\sqrt n)^k.$$
Conisder the $k=2$ term in the sum.
A: It is equivalent to prove 
$$ n < \left(1 + \frac{1}{\sqrt{n}}\right)^{2n}.$$
By the binomial identity, the right hand side is at least
\begin{align*}
& 1 + \binom{2n}{1}\frac{1}{\sqrt{n}} + \binom{2n}{2}\frac{1}{n} \\
= & 1 + 2\sqrt{n} + 2n - 1 \\
= & 2n + 2\sqrt{n} \\
> & n.
\end{align*}
A: $(1+x)^n \ge (n^2/4) x^2 $, $n \ge 1$, $x \ge 0$.
Proof: 
$(1+x)^n=$
$1+nx + (n(n-1)/2!)x^2+...\gt (n(n-1)/2)x^2 \ge (n^2/4)x^2,$
since we have 
$(n)((n-1)/2) \ge n( n/4)$,  for $n\ge 2.$
Let $x=2/√n$, in
$(1+x)^n > (n^2/4)x^2$;
$1+2/√n \gt \sqrt[n]{n}$, $n \ge 1$.
Finally 
$(1+2/√n)^2 >(1+2/√n) \gt \sqrt[n]{n}$, $n \ge 1$.
A: $$\begin{align}
\left(1+{1\over\sqrt n}\right)^{2n}
&=\left(1+{2\over\sqrt n}+{1\over n}\right)^n\\
&\gt\left(1+{2\over\sqrt n} \right)^n\\
&\ge1+{n\choose1}{2\over\sqrt n}+{n\choose2}\left(2\over\sqrt n\right)^2\\
&\gt1+0+2(n-1)\\
&=2n-1\\
&\ge n
\end{align}$$
A: Well for natural (positive) $n$ then  $\sqrt[n]{n} \le (1+ \frac 1{\sqrt n})^2 \iff$
$n \le (1 + \frac 1{\sqrt n})^{2n}$
So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n}$ and dang... that's not enough.
But lets go one more term.
Remember the reason $(1 + b)^n \ge 1+ nb$ is becase $(1 + b)^n = 1 + nb + C_2b^2 + C_3b^3 + ..... + b^n$ and $C_kb^k \ge 0$.  (Note if $b$ is large and $n>0$, in particular if $b \ge 1$, this is $(1+b)^n \ge 1 + nb$ is a very large discrepancy in the inequality).
So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n} + {2n \choose 2}\frac 1{n}$.
So ${2n \choose 2} = \frac {2n(2n-1)}2 = n(2n-1)$ so
$(1 + \frac 1{\sqrt n})^{2n} \ge 1 +\frac{2n}{\sqrt n} + n(2n-1)\frac 1n =$
$1 + 2\sqrt n + 2n- 1 = 2(\sqrt n + n) > n$.
.... which if we did that correctly is not even close.
