Prove that $a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}$

Question

Given $a,b,c>0$ such that $$a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Prove that $$a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}.$$

Answer 1

From the hypothesis, we have $$abc(a+b+c)\ge ab+bc+ca.$$ Since $$(ab+bc+ca)^2\ge 3abc(a+b+c)$$ we have $$abc(a+b+c)\ge ab+bc+ca\ge 3.$$ That means $$a+b+c\ge \frac3{abc}.$$ It remains to show that $$a+b+c\ge \frac9{a+b+c},$$ which follows from the obvious $$\frac1a+\frac1b+\frac1c\ge \frac9{a+b+c}.$$