<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] Your inequality should be not strong. 
We can prove that for all natural $n\geq1$ the following inequality holds:
$$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq\sqrt{n}\sqrt[4]{2n-1}.$$
Indeed, 
$$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq1+\int\limits_1^{n}\frac{1}{\sqrt x}dx=2\sqrt{n}-1.$$
Thus, it's enough to prove that
$$2\sqrt{n}-1\leq\sqrt{n}\sqrt[4]{2n-1}.$$
Let $n=(x+1)^2,$ where $x\geq0$.
Thus, we need to prove that
$$2x+1\leq (x+1)\sqrt[4]{2x^2+4x+1}$$ or
$$(x+1)^4(2x^2+4x+1)\geq(2x+1)^4$$ or
$$x^3(2x^3+12x^2+13x+4)\geq0,$$ which is obvious.
Done!
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