[不等式] a=b=c and 4a=2b=c

If are positive real numbers, then
\[ \frac{a(a-b)}{17a^2+4ab+6b^2}+\frac{b(b-c)}{17b^2+4bc+6c^2}+\frac{c(c-a)}{17c^2+4ca+6a^2}\geq 0, \]
with equality holds if $a=b=c$ and only if$4a=2b=c$ or or any cyclic permutation.

不知道这题能否用我介绍过的CYH技术。