# 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.

• possibly work: multiply both numerator and denominator by $a,b,c,d$ for each term, then use this inequality: $\sum\limits_{i=1}^n \frac {a_i^2}{b_i} \geq \frac{(\sum\limits_{i=1}^n a_i)^2}{\sum\limits_{i=1}^n b_i}$ – SiXUlm Jan 26 '16 at 15:44

## 2 Answers

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*}

• How is $\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd} \ge \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}$ – user19405892 Jan 26 '16 at 15:52
• @user19405892, by Cauchy-Schwartz Inequality. – user249332 Jan 26 '16 at 15:54
• @SubhadeepDey I don't see how. What are the $x_i's$ and $y_i's$? – user19405892 Jan 26 '16 at 16:01
• If I answer your question then, the comments zone will fall short. Should I Post the solution of your query in the answer zone? If you say, I can do so. – user249332 Jan 26 '16 at 16:05
• @user19405892, ok, Gordon has posted that in his answer, see this. – user249332 Jan 26 '16 at 16:08

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!

• I don't get $\sum\limits_{cyc}\frac{a}{b+2c+3d}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(4ab+2ac)}$. – user19405892 Jan 26 '16 at 15:51
• This is essentially the same solution as Gordon's. – Ewan Delanoy Jan 26 '16 at 16:11
• @MichaelRozenberg Why is the last inequality $\sum\limits_{sym}(a-b)^2\geq0$? – user19405892 Jan 26 '16 at 16:20
• @user19405892 because $\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(4ab+2ac)}\geq\frac{3}{2}\Leftrightarrow$ $3(a+b+c+d)^2\geq8(ab+ac+ad+bc+bd+cd)\Leftrightarrow$ $3(a^2+b^2+c^2+d^2)-2(ab+ac+ad+bc+bd+cd)\geq0\Leftrightarrow\sum\limits_{sym}(a-b)^2\geq0$. – Michael Rozenberg Jan 26 '16 at 18:44