Prove homogeneous inequality Prove that for all $a,b,c>0$ we have $$\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2}\leq\frac{1}{4}+\frac{4abc}{(a+b)(b+c)(c+a)}.$$
Please help me prove this homogeneous inequality.

Let 
$$
D=\frac{1}{4}+\frac{4abc}{(a+b)(b+c)(c+a)}-
\Bigg(\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2}\Bigg) \tag{1}
$$
We must show that $D$ is nonnegative. It will suffice to show
that 
$$
D=\frac{1}{4}.\frac{(b-a)^2(c-a)^2(c-b)^2}{(a+b)^2(b+c)^2(c+a)^2} \tag{2}
$$
But this follows easily enough from the expanding the identity
$$
(b-a)^2(c-a)^2(c-b)^2=((b+a)^2-4ab)((c+a)^2-4ac)((c+b)^2-4bc)
$$
(I do not put in all the details as this is homework).
