How prove this $100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$ 
let $a,b,c\ge 0$, and such $$a+b+c=6$$
show that
$$100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$$

My idea: since
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=36-2(ab+bc+ac)$$
$$a^2b^2+b^2c^2+a^2c^2=(ab+ca+bc)^2-2abc(a+b+c)=(ab+bc+ac)^2-12abc$$
so let $p=a+b+c=6, q=ab+bc+ac,r=abc$,so
$$\Longleftrightarrow  100+5[36-2q]-2[q^2-12r]-r^2\ge 0$$
$$\Longleftrightarrow  r^2+2q^2-24r+10q\le 280$$
then I can't.Thank you

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $w^3$ and we need to prove that $f(w^3)\geq0$,
where $f$ is a concave function.
But the concave function gets a minimal value for an extremal value of $w^3$, 
which happens for an equality case of two variables or when maybe $w^3=0$.
Thus, it's enough to prove our inequality in the following cases.


*

*$b=a$ and $c=6-2a$, where $0\leq a\leq 3$, which gives
$$(a-2)^2(a^2+5)(7+4a-2a^2)\geq0,$$
which is obvious for $0\leq a\leq 3$;

*$c=0$ and $b=6-a$, where $0\leq a\leq6$.
In this case we get
$$140-30a-31a^2+12a^3-a^4\geq0,$$
which is true for $0\leq a\leq6$.
Done!
