TITLE: Prove that $\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \geq \frac{2}{3}.$
QUESTION [5 upvotes]: Let $a,b,c,d$ be positive real numbers. Prove that $$\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c} \geq \dfrac{2}{3}.$$

I was thinking of trying $a \geq b \geq c$ since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn't look like I can really use AM-GM or Cauchy-Schwarz.

REPLY [2 votes]: By C-S $\sum\limits_{cyc}\frac{a}{b+2c+3d}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(4ab+2ac)}\geq\frac{2}{3}$, where the last inequality it's
$\sum\limits_{sym}(a-b)^2\geq0$. Done!