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.

The strict inequality (which only holds for $n\gt1$) can be proved by induction, by showing that
$$\sqrt n\cdot(2n-1)^{1/4}+{1\over\sqrt{n+1}}\lt\sqrt{n+1}\cdot(2n+1)^{1/4}$$
for $n\ge N$ for some $N$ and then checking the base cases up to $N$.
We first rewrite the inductive inequality above as
$$\sqrt{n(n+1)}\cdot(2n-1)^{1/4}\lt(n+1)(2n+1)^{1/4}-1$$
Noting that both sides are positive for $n\ge1$, we can square to the equivalent inequality
$$n(n+1)\sqrt{2n-1}\lt(n+1)^2\sqrt{2n+1}-2(n+1)(2n+1)^{1/4}+1$$
which is certainly true if 
$$n(n+1)\sqrt{2n+1}\lt(n+1)^2\sqrt{2n+1}-2(n+1)(2n+1)^{1/4}$$
But this reduces to $2(2n+1)^{1/4}\lt\sqrt{2n+1}$, which simplifies to $16\lt2n+1$. So the inductive inequality holds for $n\ge N=8$. And as luck would have it, the base cases for $n=2$ to $8$ have been checked in Yves Daoust's answer.
