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.
 A: Assume that for any $k\geq 1$ we have $\frac{1}{k}=a_k b_k$, with $a_k$ and $b_k$ being roughly of the same magnitude and such that both $a_k$ and $b_k$ are telescopic terms. Then
$$ \sum_{k=1}^{n}\sqrt{\frac{1}{k}}\leq \sqrt{\sum_{k=1}^{n}a_k}\sqrt{\sum_{k=1}^{n}b_k}, $$
which follows from the Cauchy-Schwarz inequality, is both an accurate and simple inequality. Let us see if we manage to find such $a_k$ and $b_k$. They both have to be close to $\frac{1}{\sqrt{k}}$, and on its turn $\frac{1}{\sqrt{k}}$ is pretty close to $2\sqrt{k+1/2}-2\sqrt{k-1/2}$, which is a telescopic term. With the choice
$$ a_k=2\sqrt{k+1/2}-2\sqrt{k-1/2},\quad b_k=\frac{\sqrt{k+1/2}+\sqrt{k-1/2}}{2k} $$
$b_k$ is not telescopic, but still $b_k\leq\frac{1}{\sqrt{k}}$. So by letting $S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$ we get:
$$ S_n \leq \sqrt{\sqrt{4n+2}-\sqrt{2}}\sqrt{S_n} $$
and the resulting inequality
$$ \sum_{k=1}^{n}\frac{1}{\sqrt{k}} = H_n^{(1/2)} \leq \sqrt{4n+2}-\sqrt{2} $$
is much sharper than $\leq \sqrt{n}(2n-1)^{1/4}$.
A: 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!
A: 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.
A: 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.
A: Here is a slightly different start than that of @michael-rozenberg, avoiding integrals in case one would like to: 
Start by writing (assuming $n>1$)
$$
\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}<
\frac{1}{\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{1}}+\cdots+\frac{2}{\sqrt{n}+\sqrt{n-1}}.
$$
Multiplying by conjugates you will find that the sum on the right-hand side equals
$$
1+2\bigl((\sqrt{2}-\sqrt{1})+\cdots+(\sqrt{n}-\sqrt{n-1})\bigr)=1+2(\sqrt{n}-1).
$$
Thus, your sum is less than
$$
2\sqrt{n}-1.
$$
Then continue in the same manner as in the nice solution by @michael-rozenberg.
