Solution by Zaratustra here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=287&t=127410&

Note $\frac{1}{\sqrt{bc}}\ge \frac{2a}{a^2+bc}$ as a result of AM-GM, so it suffices to prove $\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}$. By AM-GM $\frac{1}{2}\left( \frac{a}{bc}+\frac{b}{ca}\right) \ge \frac{1}{c}$ so $\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. By AM-GM $ \frac{1}{2}\left( \frac{1}{a}+\frac{1}{b}\right) \ge \frac{1}{\sqrt{ab}}$ so summing up analogues we are done.

2a/(a^2+bc)+2b/(b^2+ac)+2c/(c^2+ab)≤a/bc+b/ac+c/ab 2a/(a^2+bc)≤a/bc => a^2≥bc b^2≥bc c^2≥cb a^2+b^2+c^2≥ab+bc+ac 2a^2+2b^2+2c^2≥2ab+2bc+2ac (a-b)^2+(a-c)^2+(b-c)^2≥0

it is equivalent to $\sum_{cyc}a^6bc+\sum_{sym}a^5b^3-\sum_{sym}a^4b^3c-\sum_{cyc}a^3b^3c^2\ge 0$ trivialized by Muirhead.