$\frac{3}{abc} \ge a + b + c$, prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c$

Question

Let $a$, $b$ and $c$ be positive numbers such that $\frac{3}{abc} \ge a + b + c$. Prove that: $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c.$$

Any way I use - I get stuck after 2 or 3 steps...

Answer 1

$$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\ge 3\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge abc(a+b+c)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=(a+b+c)^2$$

Answer 2

The condition gives $$1\geq\sqrt{\frac{(a+b+c)abc}{3}}.$$ Thus, $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\sqrt{\frac{(a+b+c)abc}{3}}=$$ $$=\sqrt{\frac{(ab+ac+bc)^2(a+b+c)}{3abc}}\geq\sqrt{\frac{3abc(a+b+c)(a+b+c)}{3abc}}=a+b+c.$$