Inequality question: If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. 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 $\frac{1}{a^2} + \frac{1}{b^2} +  \frac{1}{c^2} + 2 (\frac{1}{ab} + \frac{1}{bc} +  \frac{1}{ca} )\geq 81$

$\frac{1}{ab} + \frac{1}{bc} +  \frac{1}{ca}$ = $\frac{c}{abc} + \frac{a}{abc} +  \frac{b}{abc}$ = ${\frac{1}{abc}}$
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=${\frac{1}{3}}$
