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 1. $ab + bc + ac \ge \frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac}$ 2. $ab- \frac {1}{ab} + bc-\frac {1}{bc} + ac- \frac {1}{ac} \ge 0$ 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: $$(ab+ac+bc)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\geq(1+1+1)^2=9$$