<s>[INST] Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
 [/INST] Denote $X=\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$ and $Y=\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}$. In fact, if $X\ge 1$, then $Y\ge \sqrt{3}$. ($\sqrt{3}$ is optimal, because when $a=b=c=\frac{1}{\sqrt{3}}$, $X=1$ and $Y=\sqrt{3}$.)
By Cauchy-Schwarz inequality,
$$3(a^2+b^2+c^2)\ge (a+b+c)^2.$$
It follows that
$$Y=\frac{a^2+b^2+c^2}{abc}\ge \frac{(a+b+c)^2}{3abc}=\frac{abc}{3}  X^2\ge \frac{abc}{3}.$$
If $abc\ge 3\sqrt{3}$, we are done. Otherwise, by inequality of arithmetic and geometric means,
$$Y\ge 3(abc)^{-\frac{1}{3}}\ge \sqrt{3}.$$
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