<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] 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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