Can We Prove This Inequality Challenge IV?

[anemone]

Prove that $\dfrac{1}{\sqrt{4x}}\le\left( \dfrac{1}{2} \right)\left( \dfrac{3}{4} \right)\cdots\left( \dfrac{2x-1}{2x} \right)<\dfrac{1}{\sqrt{2x}}$.

 

[chisigma]

Prove that $\dfrac{1}{\sqrt{4x}}\le\left( \dfrac{1}{2} \right)\left( \dfrac{3}{4} \right)\cdots\left( \dfrac{2x-1}{2x} \right)<\dfrac{1}{\sqrt{2x}}$.

[sp]Setting $\displaystyle a_{n}= \prod_{k=1}^{n} (1 - \frac{1}{2\ k})$, You can verify that $a_{n}$ obeys to the difference equation...

$\displaystyle a_{n+1} - a_{n} = - \frac{a_{n}}{2\ n}, a_{1} = \frac{1}{2}\ (1)$But (1) approximates the ODE $\displaystyle y^{\ '} = - \frac{1}{2\ x}$ the solution of which is $\displaystyle y = \frac{c}{\sqrt{x}}$, where is $c= y(1)$, so that is $a_{n} \sim \frac{a_{1}}{\sqrt{n}}$ and that leads to the conclusion.[/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

Thanks for participating, chisigma!:)

Solution suggested by other:

Spoiler

Let $A=\left( \dfrac{1}{2} \right)\left( \dfrac{3}{4} \right)\cdots\left( \dfrac{2x-1}{2x} \right)$ and $B=\left( \dfrac{2}{3} \right)\left( \dfrac{4}{5} \right)\cdots\left( \dfrac{2x-2}{2x-1} \right)$

We have $AB=\dfrac{1}{2x}$.

Notice that $\dfrac{1}{2}<\dfrac{2}{3}<\dfrac{3}{4}<\cdots<\dfrac{2x-1}{2x}$, $\therefore 2A\ge B$ so $2A^2 \ge AB=\dfrac{1}{2x}$ and from here we get$A\ge \dfrac{1}{\sqrt{4x}}$.

On the other hand, we have $A<B$, hence $A^2<AB=\dfrac{1}{2x}$ and from here we get $A<\dfrac{1}{\sqrt{2x}}$.

And therefore we reach to the desired inequality.