Having problem proving a statement

Question

I have a problem proving the following for $a,b,c>0$ $$\frac{ab}{(a+b)^2} + \frac{bc}{(b+c)^2} + \frac{ac}{(a+c)^2} \le \frac{1}{5} + \frac{4abc}{(a+b)(b+c)(c+a)}.$$

Answer 1

Counter Example $a=b=c=1$ ... $\frac{3}{4} \leq \frac{1}{5} +\frac{1}{2}$ ?

Edit : Note that \begin{eqnarray*} (a^2b+b^2c+c^2a-ab^2-bc^2-ca^2)^2 \geq 0 \end{eqnarray*} can be rearranged to give \begin{eqnarray*} (a+c)^2(b+c)^2(a+b)^2+16abc(a+c)(b+c)(a+b)\geq 4( ab(a+c)^2(b+c)^2+ac(a+b)^2(b+c)^2+cb(a+c)^2(b+a)^2). \end{eqnarray*} So the question would have worked if it had been \begin{eqnarray*} \frac{ab}{(a+b)^2} + \frac{bc}{(b+c)^2} + \frac{ac}{(a+c)^2} \le \color{red}{\frac{1}{4}} + \frac{4abc}{(a+b)(b+c)(c+a)}. \end{eqnarray*}