# Showing $\left(\frac{a}{a + 2b}\right)^2 + \left(\frac{b}{b + 2c}\right)^2 + \left(\frac{c}{c + 2a}\right)^2 \geq 1/3$

I'm trying to prove the following inequality using Holder's Inequality:

$$\left(\frac{a}{a + 2b}\right)^2 + \left(\frac{b}{b + 2c}\right)^2 + \left(\frac{c}{c + 2a}\right)^2 \geq \frac{1}{3},$$

where $a, b ,c$ are positive.

I've tried a number of combinations but I'm stuck - any ideas?

## 2 Answers

With a more generalised Holder inequality, ($\sum$ representing cyclic sums):$$\sum \left(\frac{a}{a+2b} \right)^2 \cdot \sum (a+2b) \cdot \sum a(a+2b) \geqslant \left(\sum a\right)^3$$ It remains to note $\sum (a+2b) = 3\sum a$ and $\sum a(a+2b) = (\sum a)^2$

• Very nice solution! +1. – Michael Rozenberg Oct 26 '17 at 19:06
• @MichaelRozenberg Your solution was great +1, but I couldn’t resist adding this one too :) – Macavity Oct 26 '17 at 19:33
• Beautiful - I actually wrote that down beforehand but completely failed to notice that the final cyclic sum ends up being the square of a + b + c. Thanks! – robtob Oct 27 '17 at 1:47

By C-S twice we obtain: $$\sum_{cyc}\left(\frac{a}{a+2b}\right)^2=\frac{1}{3}\sum_{cyc}1^2\sum_{cyc}\left(\frac{a}{a+2b}\right)^2\geq\frac{1}{3}\left(\sum_{cyc}\frac{a}{a+2b}\right)^2=$$ $$=\frac{1}{3}\left(\sum_{cyc}\frac{a^2}{a^2+2ab}\right)^2\geq\frac{1}{3}\left(\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+2ab)}\right)^2=\frac{1}{3}.$$ Done!

• Very nice, thanks! The problem suggested an approach with the generalised inequality but this works really nicely :) – robtob Oct 27 '17 at 1:48