Question 24
(USA 3) Prove that
$$\frac a{b+2c+3d}+\frac b{c+2d+3a}+\frac c{d+2a+3b}+\frac d{a+2b+3c}\ge\frac23$$
for all positive real numbers $a,b,c,d$.
$$\frac a{b+2c+3d}+\frac b{c+2d+3a}+\frac c{d+2a+3b}+\frac d{a+2b+3c}\ge\frac23$$
for all positive real numbers $a,b,c,d$.