# Prove that a b + 2 c + 3 d </m

Prove that $\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge \frac{2}{3}.$
I was thinking of trying $a\ge b\ge c$ 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.
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lofoptiformfp
We need only note that

EDIT:
$\begin{array}{rl}\left(a+b+c+d{\right)}^{2}& =\left(\frac{a}{\sqrt{ab+2ac+3ad}}\sqrt{ab+2ac+3ad}\\ & \phantom{\rule{2em}{0ex}}+\frac{b}{\sqrt{bc+2bd+3ab}}\sqrt{bc+2bd+3ab}\\ & \phantom{\rule{2em}{0ex}}+\frac{c}{\sqrt{cd+2ac+3bc}}\sqrt{cd+2ac+3bc}\\ & \phantom{\rule{2em}{0ex}}+\frac{d}{\sqrt{ad+2bd+3cd}}\sqrt{ad+2bd+3cd}{\right)}^{2}\\ & \le \left(\frac{{a}^{2}}{ab+2ac+3ad}+\frac{{b}^{2}}{bc+2bd+3ab}+\frac{{c}^{2}}{cd+2ac+3bc}\\ & \phantom{\rule{2em}{0ex}}+\frac{{d}^{2}}{ad+2bd+3cd}\right)\left[4\left(ab+ac+ad+bc+bd+cd\right)\right].\end{array}$
That is,

Moreover,
$\begin{array}{rl}\left(a+b+c+d{\right)}^{2}& ={a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}+2\left(ab+ac+ad+bc+bd+cd\right)\\ & =\frac{1}{3}\left[\left({a}^{2}+{b}^{2}\right)+\left({a}^{2}+{c}^{2}\right)+\left({a}^{2}+{d}^{2}\right)+\left({b}^{2}+{c}^{2}\right)+\left({b}^{2}+{d}^{2}\right)\right)\\ & \phantom{\rule{2em}{0ex}}+\left({c}^{2}+{d}^{2}\right)\right]+2\left(ab+ac+ad+bc+bd+cd\right)\\ & \ge \frac{2}{3}\left(ab+ac+ad+bc+bd+cd\right)+2\left(ab+ac+ad+bc+bd+cd\right)\\ & =\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right).\end{array}$
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spockmonkey40
By C-S $\sum _{cyc}\frac{a}{b+2c+3d}\ge \frac{\left(a+b+c+d{\right)}^{2}}{\sum _{cyc}\left(4ab+2ac\right)}\ge \frac{2}{3}$, where the last inequality it's
$\sum _{sym}\left(a-b{\right)}^{2}\ge 0$. Done!