Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq  64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq  4ab$$
$$\left ( b+ c \right )^{2}\geq  4bc$$
$$\left ( c+ a \right )^{2}\geq  4ca$$
So I want to prove $abc\geq  1$. I need to the help. Thanks!

Consider the polynomial 
$$p(x) = (x-a)(x-b)(x-c) = x^3 - Ax^2 + Bx - C
\quad\text{ where }\quad
\begin{cases}
A &= a + b + c\\
B &= ab + bc + ca\\
C &= abc
\end{cases}
$$
We are given $A = 3$. By AM $\ge$ GM, we have 
$$1 \ge C \implies \sqrt{C} \ge C$$
Since all the roots of $p(x)$ is real. By Newton's inequalities, we have
$$\left(\frac{B}{3}\right)^2 \ge \left(\frac{A}{3}\right)C  \iff B^2 \ge 3AC
\implies B \ge \sqrt{3AC} = 3\sqrt{C}$$
Combine these two inequalities, we find
$$(a+b)(b+c)(c+a) = (A-c)(A-a)(A-b) = p(A) = BA - C
\ge 3(3\sqrt{C}) - \sqrt{C} = 8\sqrt{C}\\
\implies
(a+b)^2(b+c)^2(c+a)^2 \ge 64C = 64abc
$$
