<s>[INST] Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$ Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$
For $n=1$ inequality holds.
For $n=k$
$$\left(\frac{1}{k+1}+\frac{1}{k+2}+...\frac{1}{2k}\right)^2<\frac{1}{2}$$
For $n=k+1$
$$\left(\frac{1}{k+2}+\frac{1}{k+3}+...\frac{1}{2k+2}\right)^2<\frac{1}{2}$$
I don't know how to prove this. 
Is it better to use direct method or contradiction?
 [/INST] $$ a_n = H_{2n}-H_{n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} $$
is an increasing sequence, since:
$$ a_{n+1}-a_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)}>0$$
hence:
$$ \forall n\geq 1,\qquad a_n \leq \lim_{n\to +\infty}a_n = \log(2) $$
and the problem boils down just to showing that $\log^2(2)<\frac{1}{2}$.
That follows from:
$$\log(2) = \int_{0}^{1}\frac{dx}{1+x}< \sqrt{\int_{0}^{1}\frac{dx}{(1+x)^2}} = \frac{1}{\sqrt{2}}.$$
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