<s>[INST] 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!
 [/INST] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that:
$$(9uv^2-w^3)^2\geq64u^3w^3$$ or $f(w^3)\geq0$, where
$$f(w^3)=(9uv^2-w^3)^2-64u^3w^3.$$
But, $$f'(w^3)=-2(9uv^2-w^3)-64u^3<0,$$ which says that $f$ decreases.
Thus, it's enough to prove our inequality for a maximal value of $w^3$, 
which happens for equality case of two variables.
Since $f(w^3)\geq0$ is homogeneous already, it's enough to assume $b=c=1$ and we need to prove that
$$27(a+1)^4\cdot4\geq64a(a+2)^3,$$ which is
$$(a-1)^2(11a^2+34a+27)\geq0.$$
Done!
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