RagvaloD wrote: For positive is true $$\frac{3}{abc} \geq a+b+c$$Prove $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq a+b+c$$ Romania 2005 If $a + b + c\geq \frac{1}{a} +\frac{1}{b} + \frac{1}{c}$ and $a,b,c\geq 0$ then prove:$$a +b+c\geq \frac{3}{abc}$$

$(ab+ac+bc)^2 \geq 3abc(a+b+c)$ so $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2 \geq \frac{3(a+b+c)}{abc} \geq (a+b+c)^2$

RagvaloD wrote: $(ab+ac+bc)^2 \geq 3abc(a+b+c)$ so $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2 \geq \frac{3(a+b+c)}{abc} \geq (a+b+c)^2$ Good.