# 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!

• AM-GM lead $abc\leq 1$. – Takahiro Waki Apr 4 '18 at 7:13
• You may rewrite it as $[(3-a)(3-b)(3-c)]^2 \geq 64abc$, take $ln$ on both sides then write it as $\sum[2ln(3-a)-ln(a)] \geq 6ln2$. The function $f(x)=2ln(3-x)-ln(x)$ where $0< x< 3$ has one inflection point, so you can use the '$n-1$ equal value principle'. – Shashwat1337 Feb 19 at 18:03

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

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

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

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

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

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!