<s>[INST] 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 $\frac{1}{a^2} + \frac{1}{b^2} +  \frac{1}{c^2} + 2 (\frac{1}{ab} + \frac{1}{bc} +  \frac{1}{ca} )\geq 81$
 [/INST] 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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