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

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
$$\frac1a+\frac1b+\frac1c\ge \frac9{a+b+c}.$$
