Prove that $\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \geq \frac{2}{3}.$ 
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.
 A: We need only note that
\begin{align*}
& \ \frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \\
=& \ \frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd} \\
\ge& \ \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}\\
=&\ \frac{a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)}{4(ab+ac+ad+bc+bd+cd)}\\
\ge& \ \frac{\frac{2}{3}(ab+ac+ad+bc+bd+cd)+2(ab+ac+ad+bc+bd+cd)}{4(ab+ac+ad+bc+bd+cd)}\\
=&\ \frac{2}{3}.
\end{align*}
EDIT:
\begin{align*}
(a+b+c+d)^2 &= \bigg(\frac{a}{\sqrt{ab+2ac+3ad}}\sqrt{ab+2ac+3ad} \\
&\qquad+\frac{b}{\sqrt{bc+2bd+3ab}}\sqrt{bc+2bd+3ab}\\
&\qquad+\frac{c}{\sqrt{cd+2ac+3bc}}\sqrt{cd+2ac+3bc}\\
&\qquad +\frac{d}{\sqrt{ad+2bd+3cd}}\sqrt{ad+2bd+3cd}\bigg)^2\\
&\le \bigg(\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}\\
&\qquad +\frac{d^2}{ad+2bd+3cd}\bigg)\Big[ 4(ab+ac+ad+bc+bd+cd)\Big].
\end{align*}
That is,
\begin{align*}
&\ \frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd} \\
\ge& \ \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}.
\end{align*}
Moreover,
\begin{align*}
(a+b+c+d)^2 &= a^2+b^2+c^2+d^2 +2(ab+ac+ad+bc+bd+cd)\\
&=\frac{1}{3}\Big[\big(a^2+b^2\big) + \big(a^2+c^2\big) + \big(a^2+d^2\big) + \big(b^2+c^2\big)+\big(b^2+d^2)\big)\\
&\qquad+\big(c^2+d^2\big) \Big] +2(ab+ac+ad+bc+bd+cd)\\
&\ge \frac{2}{3}(ab+ac+ad+bc+bd+cd)+2(ab+ac+ad+bc+bd+cd)\\
&=\frac{8}{3}(ab+ac+ad+bc+bd+cd).
\end{align*}
A: 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!
