How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $ 
Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$


My Approach : 
I tried by applying Tchebychev's Inequality for two same sets of numbers; 
$$1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}$$ 
And got , $$\Bigl(1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}}\Bigr)^2 <n\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr) $$
Again I tried by applying Tchebychev's Inequality for another two same sets of numbers;
$$1,\frac{1}{2},...,\frac{1}{n}$$ And got, $$\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr)^2 <n\Bigl(1 + \frac{1}{2^2} +... +\frac{1}{n^2}\Bigr)$$
With these two inequities i tried solving further more, but i couldn't. So can you please help me solving this further. And if there is some other approach for this question then please answer that way too.
Thank you.

Using Cauchy Schwartz inequality with


*

*$u=(1,1,...,1)$

*$v=\left(1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}\right)$
we have
$$\vec u \cdot \vec v\le |\vec u||\vec v| \implies 1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} \le\sqrt{n}\ \cdot\left(\sqrt{1 + \frac{1}{2} +... +\frac{1}{n}}\right)$$
and thus it suffices to prove that
$$\sqrt{n}\ \cdot\left(\sqrt{1 + \frac{1}{2} +... +\frac{1}{n}}\right)<\sqrt{n}\ .\left(2n-1\right)^{\frac{1}{4}}$$
that is
$$1 + \frac{1}{2} +... +\frac{1}{n}<\sqrt{2n-1}$$
which is true for $n=2$ and it reduces to prove by induction 
$$\sqrt{2n-1}+\frac{1}{n+1}<\sqrt{2n+1}$$
which is true indeed
$$\frac{1}{n+1}<\sqrt{2n+1}-\sqrt{2n-1}$$
$$\frac{1}{(n+1)^2}<2n+1+2n-1-2\sqrt{4n^2-1}$$
$$4n-\frac{1}{(n+1)^2}>2\sqrt{4n^2-1}$$
$$16n^2+\frac{1}{(n+1)^4}-\frac{8n}{(n+1)^2}>16n^2-4$$
$$\frac{1}{(n+1)^4}-\frac{8n}{(n+1)^2}+4>0$$
$$4(n+1)^4-8n(n+1)^2+1>0$$
$$4(n^4+4n^3+6n^2+4n+1)-8n(n^2+2n+1)+1>0$$
$$4n^4+8n^3+8n^2+8n+5>0$$
Therefore the given inequality holds for $n\ge 2$, for $n=1$ we can check directly that equality holds.
