Inequality involving four numbers Show that, if $a$, $b$, $c$ and $d$ are four positive numbers with sum $1$, then $$\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} \ge \frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}.$$ I tried to subtract the fractions, but I didn't get to any result.

Hint: It is enough to show that for $x\in (0,1)$,
$$f(x)=\frac3{1-x}-\frac5{1+x}+\frac{32}{15}(1-4x)=\frac{2(4x-1)^2(4x+1)}{15(1-x^2)}\ge0$$
