Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$

Question

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?

Answer 1

I have Cauchy-Schwarz inequality prove your inequality

we have \begin{align*}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{n+n}&\le\sqrt{n\left(\dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\cdots+\dfrac{1}{(n+n)^2}\right)}\\ &<\sqrt{n\left(\dfrac{1}{n(n+1)}+\dfrac{1}{(n+1)(n+2)}+\cdots+\dfrac{1}{(n+n-1)(n+n)}\right)}\\ &=\sqrt{n\left(\dfrac{1}{n}-\dfrac{1}{2n}\right)}\\ &=\dfrac{\sqrt{2}}{2} \end{align*}

EDIT If you must use induction, I think you must induction this following Stronger inequality
$$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{n+n}\le\dfrac{\sqrt{2}}{2}-\dfrac{1}{4n+1}$$

because we only prove this $$\dfrac{\sqrt{2}}{2}-\dfrac{1}{4n+1}-\dfrac{1}{n+1}+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}<\dfrac{\sqrt{2}}{2}-\dfrac{1}{4n+5}$$ we only prove $$\dfrac{1}{4n+5}+\dfrac{1}{2n+1}<\dfrac{1}{2n+2}+\dfrac{1}{4n+1}$$ $$\Longleftrightarrow (4n+5)(2n+1)(4n+1)+(4n+5)(2n+1)(2n+2)-(4n+5)(2n+2)(4n+1)-(2n+1)(2n+2)(4n+1)>0$$ $$\Longleftrightarrow 3>0$$ It it clear

Answer 2

$$ a_n = H_{2n}-H_{n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} $$ 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: $$\log(2) = \int_{0}^{1}\frac{dx}{1+x}< \sqrt{\int_{0}^{1}\frac{dx}{(1+x)^2}} = \frac{1}{\sqrt{2}}.$$