Let $a ; b ; c > 0$ such that $a+b+c=3$
Prove: $P=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{6abc}{ab+bc+ca} \geq 5$
I tried: $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{a^2c+b^2a+c^2b}{abc}=\dfrac{(a^2c+b^2a+c^2b)(c+a+b)}{3abc} \geq \dfrac{(ab+bc+ca)^2}{3abc}$ Then: $P \geq \dfrac{(ab+bc+ca)^2}{3abc}+\dfrac{6abc}{ab+bc+ca}$
But I think it's a wrong way, cuz $\dfrac{(ab+bc+ca)^2}{3abc}$ seems too weak.
Suppose $c = \min \{a,b,c\},$ we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} -3 = \frac{(a-b)^2}{ab} + \frac{(a-c)(b-c)}{ac},$$ $$(a+b+c)(ab+bc+ca)-9abc = 2c(a-b)^2+(a+b)(a-c)(b-c).$$ We write the inequality as $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{18abc}{(a+b+c)(ab+bc+ca)} \geqslant 5.$$ equivalent to $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} - 3 \geqslant 2\left[1 - \dfrac{9abc}{(a+b+c)(ab+bc+ca)}\right],$$ or $$\frac{(a-b)^2}{ab} + \frac{(a-c)(b-c)}{ac} \geqslant \frac{4c(a-b)^2}{(a+b+c)(ab+bc+ca)}+\frac{2(a+b)(a-c)(b-c)}{(a+b+c)(ab+bc+ca)},$$ or $$\left[\frac{1}{ab} - \frac{4c}{(a+b+c)(ab+bc+ca)}\right](a-b)^2 + \left[\frac{1}{ac} -\frac{2(a+b)}{(a+b+c)(ab+bc+ca)}\right](a-c)(b-c) \geqslant 0.$$ But $$(a+b+c)(ab+bc+ca) \geqslant 9abc \geqslant 4abc,$$ and $$\begin{aligned}(a+b+c)(ab+bc+ca) - 2ca(a+b) & \geqslant (a+b)(ab+bc+ca) - 2ca(a+b) \\&= (a+b)[a(b-c)+bc] \geqslant 0.\end{aligned}$$ The proof is completed.