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!
 A: 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
$$(a+b)(b+c)(c+a) = (A-c)(A-a)(A-b) = p(A) = BA - C
\ge 3(3\sqrt{C}) - \sqrt{C} = 8\sqrt{C}\\
\implies
(a+b)^2(b+c)^2(c+a)^2 \ge 64C = 64abc
$$
A: $$p=a+b+c=3\ , q=ab+bc+ca \ , \ r=abc\le 1 $$
$$\Leftrightarrow 3q-r \ge 8\sqrt{r}$$
$$q^2 \ge 3pr = 9r \ , \ 3q-r\ge 9\sqrt {r}-r\ge 8\sqrt {r} \Leftrightarrow r\le 1$$
A: 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!
A: Set $$f=27(a+b)^2(b+c)^2(c+a)^2-64abc(a+b+c)^3$$
Assume $a=\max \{a,b,c\}$ thus $$f=27\,(a-b)^2(b-c)^2(c-a)^2+4abc\left[11(a^3+b^3+c^3-3abc)+6\sum_{cyc} c(a-b)^2\right]+108\left\{a^2\Big[ab+c(a-b)\Big](b-c)^2+b^2c^2(a-c)(a-b)\right\} \ge 0$$
A: 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$:$ $$\text{P} =\sum\limits_{cyc}c \left( 44\,a{b}^{2}+19\,abc+49\,{c}^{2}a+5\,b{c}^{2}+27\,{c}^{3}
 \right)  \left( a-b \right) ^{2} \geqq 0$$
