<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 $$\text{P}=27\, \left( a+b \right) ^{2} \left( b+c \right) ^{2} \left( c+a
 \right) ^{2}-64\,abc \left( a+b+c \right) ^{3}$$
We have$:$ $$\text{P} =\sum\limits_{cyc}c \left( 44\,a{b}^{2}+19\,abc+49\,{c}^{2}a+5\,b{c}^{2}+27\,{c}^{3}
 \right)  \left( a-b \right) ^{2} \geqq 0$$
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