<s>[INST] 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


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

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


But now I don't know what to do. 
 [/INST] from the condition we have $$a+b+c=3abc$$ and by AM-GM we get $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$ plugging the condition in this inequlity we get $$abc\geq1$$ we will prove the inequality:
By AM-GM we have $$ab+bc+ca\geq 3\sqrt[3]{(abc)^2}\geq 3 $$ if $$abc\geq 1$$
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