If $abc=1$ so $\sum\limits_{cyc}\frac{a^3c}{(a+c)(b+c)}\geq\frac{3}{4}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a^3c}{(a+c)(b+c)}+\frac{b^3a}{(b+a)(c+a)}+\frac{c^3b}{(c+b)(a+b)}\geq\frac{3}{4}$$
I tried C-S and BW. It does not help. 
 A: Let $a = \frac{x}{y}, \ b = \frac{y}{z}, \ c = \frac{z}{x}; \ x, y, z > 0$.
It suffices to prove that $f(x, y, z)\ge 0$ where
\begin{align*}
f(x,y, z) &= 4\, x^7\, z^5 - 3\, x^6\, y^3\, z^3 + 4\, x^6\, y^2\, z^4 + 4\, x^5\, y^7 - 3\, x^5\, y^5\, z^2 - 3\, x^5\, y^2\, z^5 + 4\, x^4\, y^6\, z^2 - 6\, x^4\, y^4\, z^4\\
 &\quad - 3\, x^3\, y^6\, z^3 - 3\, x^3\, y^3\, z^6 - 3\, x^2\, y^5\, z^5 + 4\, x^2\, y^4\, z^6 + 4\, y^5\, z^7.
\end{align*}
WLOG, assume that $z = \min(x,y,z).$ There are two possible cases:
1) $z \le y \le x$: Let $z = 1, \ y = 1+s, \ x = 1+s + t; \ s,t \ge 0$. We have
\begin{align*}
&f(1+s+t, 1+s, 1)\\
 =\ & 4\, t^7 + \left( - 3\, s^3 - 5\, s^2 + 27\, s + 29\right)\, t^6 + \left(4\, s^7 + 28\, s^6 + 81\, s^5 + 107\, s^4 + 62\, s^3 + 99\, s^2 + 175\, s + 88\right)\, t^5 \\
  &\quad + g(s,t)\\
  \ge \ & 0 ,
 \end{align*}
since 
\begin{align*}
&4\cdot 4 \cdot (4\, s^7 + 28\, s^6 + 81\, s^5 + 107\, s^4 + 62\, s^3 + 99\, s^2 + 175\, s + 88) -
( - 3\, s^3 - 5\, s^2 + 27\, s + 29)^2\\
 =\ & {\left(s + 1\right)}^2\, \left(64\, s^5 + 311\, s^4 + 580\, s^3 + 378\, s^2 + 100\, s + 567\right)\\
 \ge \ & 0,
\end{align*} where $g(s,t)$ is a polynomial with non-negative coefficients.
2) $z\le x \le y$: Let $z = 1, \ x = 1+s, \ y = 1+s+t; \ s,t\ge 0.$
$f(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients. The inequality is true. This completes the proof.
