prove $ b^6c^8-b^5c^7+b^8c^6-b^3c^6+b^2c^6-b^7c^5-b^6c^3+b^6c^2-bc^2+c^2-b^2c+b^2 \ge 0$ $b>0,c>0, $  prove $g(b,c)=b^6c^8-b^5c^7+b^8c^6-b^3c^6+b^2c^6-b^7c^5-b^6c^3+b^6c^2-bc^2+c^2-b^2c+b^2 \ge 0$
this is from a middle process of a inequality. (I am sure it is correct because the inequality is proved.)
edit: the inequality is : $abc=1,\dfrac{a}{b^2+c^2+a}+\dfrac{b}{c^2+a^2+b}+\dfrac{c}{a^2+b^2+c} \le 1$ , replace $a=\dfrac{1}{bc}$ ,one can get the  asked inequality.
I try different approach such as $c=b+u,$  or $ c=tb \to b^{12}t^8-b^{10}t^7+b^{12}t^6-b^7t^6+b^6t^6-b^{10}t^5-b^7t^3+b^6t^2-bt^2+t^2-bt+1 \ge 0$
it doesn't work.there is no factor such as $(t-1),(b-1),(bt-1),(b^2t-1),(b-c)$
when $c=b$ we have $(b-1)^2f(b)$ and $ f(b) >0$  so it is clear that $g(b,c)=0 \iff b=c=1$, but how to prove it?  
thanks in advance.

It seems the following. 
AM-GM inequality implies the following inequalities: 
$$\frac 12 b^8c^6+\frac 16 b^6c^8+\frac 13 b^6c^2\ge b^7c^5,$$
$$\frac 12 b^6c^8+\frac 16 b^8c^6+\frac 13 b^2c^6\ge b^5c^7,$$
$$\frac 13 b^8c^6+\frac 12 b^6c^2+\frac 16 b^2\ge b^6c^3,$$
$$\frac 13 b^6c^8+\frac 12 b^2c^6+\frac 16 c^2\ge b^3c^6,$$
$$\frac 16 b^6c^2+\frac 56 c^2\ge bc^2,$$
$$\frac 16 b^2c^6+\frac 56 b^2\ge b^2c.$$
When we add all of them, we obtain $g(b,c)\ge 0$.
