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.

• Is it impolite to notice that the inequality is obvious when $n\geqslant4$ since, for every $n$, $\frac{2n}{n+1}<2$ and $1+\frac12+\frac13+\frac14>2$, and that checking it for $n=1,2,3$ takes about 5 seconds? Hence that this must be one of the most unconvincing applications of Cauchy-Schwarz one can imagine...
– Did
Oct 26, 2015 at 21:06

2 Answers

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 $<u,v>$, $\|u\|$ and $\|v\|$. Recall that $\displaystyle 1+2+\cdots+n=\frac{n(n+1)}{2}$.

• I'm having trouble understanding how that relates to the given inequality. Oct 26, 2015 at 20:43
• The Cauchy-Schwarz inequality say that $|<u,v>|\leq \|u\| \|v\|$. Have you computed these quantities ? Oct 26, 2015 at 20:51
• What I'm getting is $||u|| = \sqrt{\sum{\frac{1}{n}}}$, $||v|| = \sqrt{\sum{n}} = \frac{n(n+1)}{2}$, $<u,v> = \sqrt{\sum{1}} = n$. Oct 26, 2015 at 21:09
• For $\|v\|$ this is $\sqrt{\frac{n(n+1)}{2}}$, for $<u,v>=1+1+\cdots+1=n$ (and $\|u\|=\sqrt{1+\frac{1}{2}+\cdots+\frac{1}{n}}$ ) Oct 26, 2015 at 21:12
• Ahh, yes that makes sense now. So it would go $n \leq \sqrt{\sum{\frac{1}{n}}}\sqrt{\frac{n(n+1)}{2}}$. Squaring both sides we get $n^2 \leq \frac{n(n+1)}{2}\sum{\frac{1}{n}}$. Solving we're left with $\frac{2n}{n+1} \leq \sum{\frac{1}{n}}$ Oct 26, 2015 at 21:20

$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\geq \frac{(1+1+1+...+1)^2}{1+2+3+...+n}=\frac{2n^2}{n(n+1)}=\frac{2n}{n+1}$