Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$

Question

Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$ for $a, b, c \in \mathbb{R_{>0}}$

I tried by first using AM-HM inequality on $a, b, c$ to get the following result.

$\frac{a+b+c}3 \geq \frac 3{\frac 1a+\frac1b+\frac1c}$

$\implies (a+b+c)(\frac1a+\frac1b+\frac1c) \geq 9$

$\implies (a+b+c)(ab+bc+ca) \geq 9abc$

Also I used the inequality

$(a+b)(b+c)(c+a) \geq 8abc$

But then I am not able to proceed further. Can someone please help me.

Answer 1

$$9(a+b)(b+c)(c+a) - 8(a+b+c)(ab+bc+ca)$$ $$ = (a+b)(b+c)(c+a) - 8abc$$ $$\ge0$$

because $(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca)-abc$

and $(a+b)(b+c)(c+a) \ge 8abc$ by AM-GM: $(a+b)(b+c)(c+a)=abc+\dots+bca\ge 8\sqrt[8]{a^8b^8c^8}$.

Answer 2

Before I start my solution, I want to say that the proof you have just showed was nice try but it's hard to prove the claim. If you want to show that $A>B$, it's meaningless to show that $A>C$ and $B>C$.

Alright, so I'll show my solution.

\begin{align} & \text{Claim. } 9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca). \\ & \text{pf)} \\ \ \\ \text{Claim} & \Leftrightarrow 9\Bigg(\sum_{sym} a^2b + 2abc \Bigg) \geq 8 \Bigg( \sum_{sym} a^2b + 3abc \Bigg) \\ & \Leftrightarrow \sum_{sym} a^2b \geq 6abc \\ & \Leftrightarrow a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \geq 6abc \\ & \Leftrightarrow \dfrac {a^2b+b^2c+c^2a+ab^2+bc^2+ca^2} {6} \geq \sqrt[6]{a^2b \cdot b^2c \cdot c^2a \cdot ab^2 \cdot bc^2 \cdot ca^2} \\ & \Rightarrow \text{Proved by AM-GM Inequality.} \end{align}