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!

Set $$f=27(a+b)^2(b+c)^2(c+a)^2-64abc(a+b+c)^3$$
Assume $a=\max \{a,b,c\}$ thus $$f=27\,(a-b)^2(b-c)^2(c-a)^2+4abc\left[11(a^3+b^3+c^3-3abc)+6\sum_{cyc} c(a-b)^2\right]+108\left\{a^2\Big[ab+c(a-b)\Big](b-c)^2+b^2c^2(a-c)(a-b)\right\} \ge 0$$
