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If $a,b,c > 0$ are such that $abc=1$, then $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c. $$

I would be pleased if you give me a hint. Thanks in advance.

Droplets
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  • $ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{a^2c+b^2a+bc^2}{abc} = a^2c+b^2a+bc^2$...can you take it from here? – DanielY Oct 31 '13 at 08:51

1 Answers1

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$$a+b+c =$$

$$\{ \mbox{multiply by 1: maybe } abc\mbox{, maybe }\sqrt{abc}, ... \}$$

$$= \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}=$$

$$ =\sqrt[3]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c}} +\sqrt[3]{\frac{b}{c} \cdot \frac{b}{c} \cdot \frac{c}{a}} +\sqrt[3]{\frac{c}{a} \cdot \frac{c}{a} \cdot \frac{a}{b}} \le $$ $$ \{\mbox{ GM}\le\mbox{AM }\} $$ $$ \le \frac{\frac{a}{b}+\frac{a}{b}+\frac{b}{c}}{3} + \frac{\frac{b}{c}+\frac{b}{c}+\frac{c}{a}}{3} + \frac{\frac{c}{a}+\frac{c}{a}+\frac{a}{b}}{3} = $$ $$ =\frac{a}{b}+\frac{b}{c}+\frac{c}{a}. $$

Oleg567
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    Hi! I have always been fascinated by these seemingly contrived algebraic manipulations. Can I ask how you do them? Do you just try random manipulations until you recognize something? Or start with a goal of which inequality to use? – Ovi Jun 02 '18 at 03:01
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    Hi @Ovi! First step was to kill dimensions. Imagine that $a,b,c$ are measured in 'meters'. So, LHS is dimension-free, and RHS is $a+b+c; (m)$; and we can imagine that $abc=1(m^3)$. Therefore we divide RHS by $\sqrt[3]{abc} (meters)$ to "equalize" dimensions.

    Then when you'll see $\sqrt[3]{\ldots}$, the most intuitive second step is to apply AM-GM. ...

     – Oleg567 Jun 03 '18 at 10:56
  • ... Few links which are my lovely ones of this sort:

    https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means,

    https://en.wikipedia.org/wiki/Generalized_mean,

    https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality,

    https://en.wikipedia.org/wiki/Rearrangement_inequality;

    here are few examples which can develop intuition in this branch:

    https://www.cut-the-knot.org/m/Algebra/CyclicInequality1.shtml .

     – Oleg567 Jun 03 '18 at 10:57