Prove the inequality, fractions. $\{ a,b,c \in\Bbb R_+\ \}$ If $\frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac} = 3$ then prove the inequality:
$ab + bc + ac \ge 3$
How I started


*

*$ab + bc + ac \ge \frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac}$

*$ab- \frac {1}{ab} + bc-\frac {1}{bc} + ac- \frac {1}{ac} \ge 0$


But now I don't know what to do. 

We need to prove that
$$ab+ac+bc\geq\frac{9}{\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}}$$ or
$$(ab+ac+bc)(a+b+c)\geq9abc$$
which is just AM-GM:
$$(ab+ac+bc)(a+b+c)\geq3\sqrt[3]{a^2b^2c^2}\cdot3\sqrt[3]{abc}=9abc$$
Or by C-S:
$$(ab+ac+bc)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\geq(1+1+1)^2=9$$
