Hint

The inequality can be written as: $\frac{a^2+b^2+c^2}{abc}\geq\frac{2bc+2ca-2ab}{abc}$ $<=>(a+b)^2-2c(b+a)+c^2\geq0$ $<=>(a+b-c)^2\geq0$

The hint actually gives way to the answer...

Olympus_mountaineer wrote: Prove that,if $a,b,c$ are positive real numbers, \[ \dfrac{a}{bc}+ \dfrac{b}{ca}+\dfrac{c}{ab}\leq \dfrac{2}{a}+\dfrac{2}{b}-\dfrac{2}{c}\] Maybe.\[ \dfrac{a}{bc}+ \dfrac{b}{ca}+\dfrac{c}{ab}\geq \dfrac{2}{a}+\dfrac{2}{b}-\dfrac{2}{c}\]

Yes,I fixed the problem.

https://artofproblemsolving.com/community/c6h53823p336839

Steve12345 wrote: https://artofproblemsolving.com/community/c6h53823p336839 I am surprized that it is a well known inequality!

use AM-GM inequality

$(a+b-c)^2\ge0\Rightarrow a^2+b^2+c^2\ge2ac+2bc-2ab\Rightarrow\frac a{bc}+\frac b{ca}+\frac c{ab}\ge\frac2a+\frac2b-\frac2c$ It holds for all $a,b,c$ with $a^2+b^2+c^2\ne0$.