Looking to confirm inequality or learn where mistake is Hello I am looking for some advice on the following,
I am wanting to show that $n^{\frac{1}{n}} \lt (1+\frac{1}{\sqrt{n}})^{2}$ for all $n \in \mathbb{N}$
and I thought I would try by induction 
The base case of n=1 is clear because $1 \lt 4$
Now I said suppose it holds that
$$n^{\frac{1}{n}} \lt (1+\frac{1}{\sqrt{n}})^{2}$$
then I must show this implies the truth of
$$(n+1)^{\frac{1}{n+1}} \lt (1+\frac{1}{\sqrt{n+1}})^{2}$$
I used Bernoulli to show that $(1+\frac{1}{\sqrt{n+1}}^{n+1} \gt \sqrt{n+1}$ for all n and then I thought if I can just use that and show that $\sqrt{n+1} \gt (n+1)^{\frac{1}{n+1}}$ then my proof would be complete.
Any advice guys? How does it look? I tried to do as much as I can on my own as I do want to learn, but I also dont want to be coming up with false proofs etc
Here is my new way of saying it,
We already showed base case,
now suppose that $$n^{1/n} \lt (1+\frac{1}{\sqrt{n}})^{2}$$
Then we want to show   $$(n+1)^{1/(n+1)} \lt (1+\frac{1}{\sqrt{n+1}})^{2}$$
I then use $$(1+\frac{1}{\sqrt{n+1}})^{2} \ge 1+2\sqrt{n+1} \gt \sqrt{n+1} \gt (n+1)^{\frac{1}{n+1}}$$ to show it holds,
how does this seem?
 A: Applying the AM-GM inequality:
$\sqrt[n]{n} = \sqrt[n]{1\cdot 1\cdot 1\cdots 1\cdot \sqrt{n}\cdot \sqrt{n}}\leq \dfrac{n-2+2\sqrt{n}}{n}<1+\dfrac{2}{\sqrt{n}}<1+\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}=\left(1+\dfrac{1}{\sqrt{n}}\right)^2$
A: Your approach is fine, except you have not written out the steps cleanly.  As I understand, you want to use induction and hence the inductive step
$$\left(1+\frac1{\sqrt{n+1}}\right)^2 > (n+1)^{\frac1{n+1}} \iff \left(1+\frac1{\sqrt{n+1}}\right)^{n+1} > \sqrt{n+1}$$
Where the last step can be proved using Bernoulli's inequality.

Of course it is simpler to use Bernoulli directly and avoid induction altogether.  The original inequality, by raising both sides to the power $\dfrac{n}2$,  is equivalent to
$$\left(1+\frac1{\sqrt{n}}\right)^n> \sqrt{n}$$
and that is easily concluded by Bernoulli which in fact gives the tighter $LHS > 1+\sqrt{n}$.
A: Hint: Equivalently we want to show that $n\lt \left(1+\frac{1}{\sqrt{n}}\right)^{2n}$.
For $n\ge 2$, use the first three terms of the binomial expansion of $\left(1+\frac{1}{\sqrt{n}}\right)^{2n}$.
