<s>[INST] 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.
 [/INST] 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.
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