<s>[INST] 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?
 [/INST] 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!
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