Prove Inequality with Cauchy Schwarz

Use the Cauchy-Schwarz Inequality to show that for any positive integer n, $\frac{2n}{n+1} \leq 1 + \frac{1}{2} + \cdots + \frac{1}{n}$

I'm having some trouble understanding how the Cauchy Schwarz Inequality can be applied to this. I've tried separating the $\frac{2n}{n+1}$ into two parts, but I'm getting nowhere with that.

Answer

Hint: Take as vectors in $\mathbb{R}^n$ $\displaystyle u=(1,\frac{1}{\sqrt{2}},\cdots,\frac{1}{\sqrt{n}})$, $\displaystyle v=(1,\sqrt{2},\cdots, \sqrt{n})$, and compute $$,$\|u\|$and$\|v\|$. Recall that$\displaystyle 1+2+\cdots+n=\frac{n(n+1)}{2}$.