amparvardi wrote: Let $a,b,c,d$ be positive real numbers. Prove that \[\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \geq \frac{2}{3}.\] Obvious from the Cauchy-Schwarz inequality applied as follows: \[\sum_{cyc}\frac{a}{b+2c+3d}=\sum_{cyc}\frac{a^2}{ab+2ca+3ad}\geq \frac{(a+b+c+d)^2}{\sum (ab+2ca+3ad)}.\] The rest that we have to show is \[\frac 32(a+b+c+d)^2\geq 4\sum_{sym}ab;\] Which is equivalent to showing that \[3(a^2+b^2+c^2+d^2)\geq 2\sum_{sym}ab,\] Which is a direct consequence of the AM-GM Inequality. $\Box$

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