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Let $a,b,c$ be three real positive numbers such that $abc\geq 1$. Prove that $$\frac{a}{b} + \frac{b}{c} +\frac{c}{a} ≥ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$

Here's what I have done. The expression is equivalent to $$a^2c+b^2a+c^2b \geq ab+bc+ac .$$ Denote $f(a,b,c)=a^2c+b^2a+c^2b-(ab+bc+ac)$. We need to show that $f(a,b,c)\geq 0$.

By AM-GM, $a^2c+b^2a+c^2b \geq 3$ (Given, $abc\geq 1$). By AM-HM, $(a+b+c)(1/a +1/b +1/c)\geq 9$ by AM-GM, min value of $a+b+c=3$ so $1/a + 1/b + 1/c \geq 9/3=3$. Thus, The minimal value of $f(a,b,c)$ is equal to the min value of $a^2c+b^2a+c^2b$ minus the min value of $ab+bc+ca=3-3=0$.

Please let me know if there is fault in argument.

Michael Rozenberg
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Epsilon zero
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  • Please format your questions so that they are more readable. You may want to have a look here: https://math.meta.stackexchange.com/questions/12660/how-to-do-formatting – uniquesolution Jun 20 '17 at 08:26
  • There is a similar problem here: https://math.stackexchange.com/questions/2054232 but it's just another problem. The starting problem is not duplicate! – Michael Rozenberg Jun 20 '17 at 12:53
  • @MichaelRozenberg: Well, the difference is $abc \ge 1$ here vs $abc = 1$ in the other question (so the other inequality is sharper). But the same solutions apply, and actually your answer https://math.stackexchange.com/a/2054313/42969 is almost identical to your answer below. – Martin R Jun 20 '17 at 14:20
  • Dear @Martin In Math even very little "almost" can give a big problem. An inequality with condition like $abc\geq1$ can be much harder than inequality with condition $abc=1$. In our case it's not so harder, but it's harder. Id est, these problems are just similar problems only. The starting problem is not duplicate. It's my opinion of course, but I think all human, which like to prove inequalities, will say that I am right. – Michael Rozenberg Jun 20 '17 at 14:42

1 Answers1

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By AM-GM we obtain: $$\sum_{cyc}\frac{a}{b}=\frac{1}{3}\sum_{cyc}\left(\frac{a}{b}+\frac{2b}{c}\right)\geq\sum_{cyc}\sqrt[3]{\frac{ab}{c^2}}=\sum_{cyc}\sqrt[3]{\frac{abc}{c^3}}\geq\sum_{cyc}\frac{1}{a}.$$ Done!

Michael Rozenberg
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