<s>[INST] Prove that $3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) \ge 10 + 8\cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$ 
For the positive real numbers $a, b, c$ prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$$

I did the following:
$$\begin{split}\dfrac{a^2+b^2+c^2}{ab+bc+ca} & = \dfrac{(a+b+c)^2-2(ab+bc+ca)}{ab+bc+ca} \\ & = \dfrac{(a+b+c)^2}{ab+bc+ca} - 2 \\ & \le \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} -2\end{split}$$
I got the last one by applying Titu's Lemma
Thus I think it suffices to prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\bigg(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-2\bigg) \\ \Longrightarrow 3a^2b+3b^2c+3c^2a+6abc \ge 5a^2c+5b^2a+5c^2b$$
But by rearrangement inequality this is not always true I think. The direction of the inequality should have been flipped.
 [/INST] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, your inequality is a linear inequality of $w^3$.
Hence, it remains to prove it for an extremal value of $w^3$, which happens in the following cases.


*

*$w^3\rightarrow0^+$. In this case our inequality is obvious;

*$b=c=1$, which gives $(a-1)^2(2a+3)\geq0$. Done!
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