# Analysis Inequality

1. Oct 1, 2008

### hypermonkey2

hey guys, i came across this inequality in analysis and am not sure how to prove it. Any ideas? It's not homework, im just curious..

Let a1, a2, . . . , an be strictly positive real numbers. Show that

a1 + a2 + · · · + an−1 + an <= ((a1)^2)/a2 + ((a2)^2)/a3) +...+ ((an)^2)/a1

cheers

2. Oct 1, 2008

### morphism

Here's the case n=2. It follows from the AM-GM inequality.

$$0 \leq \left(\frac{a_1}{\sqrt{a_2}} - \frac{a_2}{\sqrt{a_1}}\right)^2 = \frac{a_1^2}{a_2} - 2\sqrt{a_1 a_2} + \frac{a_2^2}{a_1} \leq \frac{a_1^2}{a_2} - (a_1 + a_2) + \frac{a_2^2}{a_1}.$$

3. Oct 1, 2008

### hypermonkey2

Thats pretty sweet. Thats hard to generalize, no?

4. Oct 1, 2008

### morphism

Yeah. I thought it was an easy induction argument after that, but that's obviously not the case. The flu is making me stupid.

Anyway, we can use the Cauchy-Schwarz inequality instead.

$$\left(\sum a_i\right)^2 = \left(\sum \frac{a_i}{\sqrt{a_{i+1}}} \sqrt{a_{i+1}}\right)^2 \leq \sum \frac{a_i^2}{a_{i+1}} \sum a_{i+1},$$

where the index of summation is taken mod n (so a_{n+1} = a_1).