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$\{ a,b,c \in\Bbb R_+\ \}$ 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.

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    Harmonic mean $\le$ arithmetic mean. – 2017-02-21
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    hello Martin, we don't need that only $AM-GM$ is needing – 2017-02-21
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    Well, you could also say that you don't "need" AM-GM because it follows from the rearrangement inequality. –  I consider $HM \le GM \le AM$ "well-known" and it directly solves this problem. – 2017-02-21

2 Answers 2

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

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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 $$abc\geq 1$$