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

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

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

• Commented Apr 14, 2022 at 7:05
• I believe, you've missed something in problem statement. Check case $a=-1$, $b=-1$, $c=0$. Commented Apr 14, 2022 at 9:09
• While we use a.m.-g.m.-inequality we always consider $a, b, c$ for $\mathbb{R_{> 0}}$ Commented Apr 14, 2022 at 12:34

$$9(a+b)(b+c)(c+a) - 8(a+b+c)(ab+bc+ca)$$ $$= (a+b)(b+c)(c+a) - 8abc$$ $$\ge0$$
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: $$(a+b)(b+c)(c+a)=abc+\dots+bca\ge 8\sqrt[8]{a^8b^8c^8}$$.
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 $$B>C$$.
\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}