Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$

Question

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!

Answer 1

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

$$(a+b)(b+c)(c+a) = (A-c)(A-a)(A-b) = p(A) = BA - C \ge 3(3\sqrt{C}) - \sqrt{C} = 8\sqrt{C}\\ \implies (a+b)^2(b+c)^2(c+a)^2 \ge 64C = 64abc $$

Answer 2

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!

Answer 3

$$p=a+b+c=3\ , q=ab+bc+ca \ , \ r=abc\le 1 $$

$$\Leftrightarrow 3q-r \ge 8\sqrt{r}$$

$$q^2 \ge 3pr = 9r \ , \ 3q-r\ge 9\sqrt {r}-r\ge 8\sqrt {r} \Leftrightarrow r\le 1$$

Answer 4

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

Answer 5

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