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.

I’ll show that the function $$g(a,b,c,d)=\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} - \left(\frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}\right)$$ is minimized over positive $a,b,c,d$ with sum $1$ when $a=b=c=d=\frac{1}{4}$. Its minimum is then $g(\frac14,\frac14,\frac14,\frac14)=0$, proving the inequality. 
Consider the function $$f(x,y)=\frac 3 {1-x} + \frac 3 {1-y}  - \left(\frac 5 {1+x} + \frac 5 {1+y} \right).$$
Its directional derivative in the direction $(1,-1)$ is $$3\left(\frac{1}{(1-x)^2}-\frac{1}{(1-y)^2}\right)+5\left(\frac{1}{(1+x)^2}-\frac{1}{(1+y)^2}\right),$$ which is negative when $x<y$ and positive when $x>y$. Consequently, if $a<b$ and $0<\epsilon<\frac{b-a}{2}$, $$f(a+\epsilon,b-\epsilon)< f(a,b),$$ and therefore $g$ can be made smaller by averaging (or making closer) any two of its parameters that are unequal (which leaves $a+b+c+d$ unchanged).
