Prove that $\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \geq \frac{2}{3}.$

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*}

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!