Let $a,b,c$ are all positive real numbers. Prove

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

Can anyone give a hint?


By Cauchy-Schwarz,

$$\sum \limits_{\text{cyc}} \left ( \dfrac{a}{a+2b} \right )^2 \geq \dfrac{1}{3} \left(\sum \limits_{\text{cyc}} \left ( \dfrac{a}{a+2b} \right )\right)^2 = \dfrac{1}{3}\left( \sum \limits_{\text{cyc}} \left ( \dfrac{a^2}{a^2+2ab} \right) \right)^2 \geq \dfrac{1}{3} \left ( \dfrac{(a+b+c)^2}{(a+b+c)^2} \right )^2 = \dfrac{1}{3}.$$

  • $\begingroup$ Note that the asker was looking for a hint. $\endgroup$ – user Nov 3 '18 at 16:15
  • $\begingroup$ I was trying to use Cauchy-Schwarz but I approached it in a really complicated way. Yours is much better. $\endgroup$ – S.stark Nov 3 '18 at 16:42

By C-S and by the Vasc's inequality we obtain: $$\sum_{cyc}\frac{a^2}{(a+2b)^2}=\sum_{cyc}\frac{a^4}{a^2(a+2b)^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+4a^3b+4a^2b^2)}=$$ $$=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}4a^3b+\sum\limits_{cyc}2a^2b^2+(a^2+b^2+c^2)^2}\geq\tfrac{(a^2+b^2+c^2)^2}{\frac{4}{3}(a^2+b^2+c^2)^2+\frac{2}{3}(a^2+b^2+c^2)^2+(a^2+b^2+c^2)^2}=\frac{1}{3}.$$


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