<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] 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.
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