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If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. I've tried AM-HM but it gave $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9$ which gives $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + 2 (\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} )\geq 81$

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3 Answers 3

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$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$ = $\frac{c}{abc} + \frac{a}{abc} + \frac{b}{abc}$ = ${\frac{1}{abc}}$

We know that AM${\ge}$GM

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

${\frac{1}{3}}$ ${\ge} \sqrt[3]{ abc}$

${\frac{1}{27}}$ ${\ge} { abc}$

${\frac{1}{abc}}$ ${\ge} 27$

so minimum value is 27

This minimum value is achieved when a=b=c=${\frac{1}{3}}$

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Assuming $a,b,c$ are positive reals,

$$\sum\limits_{\textrm{cyc}}\frac 1{ab}=\frac 1{abc}\sum\limits_{\textrm{cyc}}a=\frac 1{abc}\geq \frac 1{\left(\frac{a+b+c}3\right)^3}=27$$

with equality when $a=b=c=1/3$ giving $3\cdot\dfrac 1{(1/3)^3}=27$

where the penultimate step follows from the AM-GM inequality

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By Holder $$\sum_{cyc}\frac{1}{ab}=\sum_{cyc}\frac{1}{ab}\sum_{cyc}a\sum_{cyc}b\geq\left(\sum_{cyc}\sqrt[3]{\frac{1}{ab}\cdot a\cdot b}\right)^3=\left(\sum_{cyc}1\right)^3=27.$$ The equality occurs for $a=b=c=\frac{1}{3}$, which says that $27$ is a minimal value.

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