<s>[INST] 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
 [/INST] Let $G(a,b,c)=100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2$. We will
show that $G(a,b,c) \geq G(2,2,2)=0$. 
From the identity
$$
4\frac{G(a,b,c)-G(a,\frac{b+c}{2},\frac{b+c}{2})}{(c-b)^2}
+\frac{G(a,b,c)-G(a,b+c,0)}{bc}=\frac{(a^2+2)(b+c)^2}{4}
$$
we see that at least one of the two numbers 
$G(a,b,c)-G(a,\frac{b+c}{2},\frac{b+c}{2})$ or
$G(a,b,c)-G(a,b+c,0)$ is nonnegative, so that it suffices to treat the cases when
one variable is zero or when two variables are equal. This is what we do in the next
two lemmas.
Lemma 1. $G(a,b+c,0) \geq G(3,3,0)=28$.
Proof of lemma 1 Putting $d=b+c=6-a$, and $x=a-3$ we have
$$
\begin{array}{lcl}
G(a,d,0) &=& 100+5(a^2+d^2)-2a^2d^2 \\
&=& 100+5((3+x)^2+(3-x)^2)-2((9-x^2)^2) \\
&=& 28+46x^2-2x^4 \geq 28=G(3,3,0).   
\end{array}
$$ 
Lemma 2. $G(a,\frac{b+c}{2},\frac{b+c}{2}) \geq G(2,2,2)=0$.
Proof of lemma 2 Putting $e=\frac{b+c}{2}=\frac{6-a}{2}$, we have
$$
\begin{array}{lcl}
G(a,e,e) &=& 100+5(a^2+2e^2)-2(2a^2e^2+e^4)-a^2e^4 \\
&=& 100+5\bigg(a^2+2\big(\frac{6-a}{2}\big)^2\bigg)
-2\bigg(2a^2\bigg(\frac{6-a}{2}\bigg)^2+\bigg(\frac{6-a}{2}\bigg)^4\bigg)
-a^2\bigg(\frac{6-a}{2}\bigg)^4 \\
&=& (a-2)^2\Bigg(7+a\bigg(\frac{7}{4}+\frac{9(6-a)}{8}+\frac{(6-a)^2}{8}+\frac{(6-a)^3}{16}\bigg)
\bigg) \Bigg) \geq 0.   
\end{array}
$$
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