Can You Prove This Inequality Involving Fractions and Square Roots?

[anemone]

Prove that $1-\dfrac{1}{2014}\left(\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2015}\right)>\dfrac{1}{\sqrt[2014]{2015}}$

---

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem.:(

You can view the proposed solution below:

Spoiler

Note that we could rewrite the LHS of the given inequality as

$\begin{align*}1-\dfrac{1}{2014}\left(\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2015}\right)&=\dfrac{1}{2014}\left(\overbrace{1+1+1+\cdots+1}^{2014}-\left(\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2015}\right)\right)\\&=\dfrac{1}{2014}\left(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)\cdots+\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)\right)\\&=\dfrac{1}{2014}\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\cdots+\dfrac{2013}{2014}+\dfrac{2014}{2015}\right)\end{align*}$

At this point, we can apply the AM-GM inequality to

$\begin{align*}\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\cdots+\dfrac{2013}{2014}+\dfrac{2014}{2015}&\ge 2014\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdots\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\right)^{\frac{1}{2014}}\\& \ge 2014\left(\frac{1}{2015}\right)^{\frac{1}{2014}}\end{align*}$

Therefore

$\begin{align*}1-\dfrac{1}{2014}\left(\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2015}\right)&=\dfrac{1}{2014}\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\cdots+\dfrac{2013}{2014}+\dfrac{2014}{2015}\right)\\&\ge \dfrac{1}{2014}\left(2014\left(\frac{1}{2015}\right)^{\frac{1}{2014}}\right)\\&\ge \left(\frac{1}{2015}\right)^{\frac{1}{2014}}\text{Q.E.D.}\end{align*}$