<s>[INST] prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$ If $a,b,c$
are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$
After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$
 [/INST] Let $x=q-r$, $y=r-p$, and $z=p-q$. 
Then $x+y+z=0$, and the problem is equivalent to 
$$a^2 yz+b^2 zx+ c^2 xy \leq 0.$$
If we substitute $x$ with $-y-z$, then the above inequality is equivalent to 
$$(a^2-b^2-c^2)yz \leq b^2 z^2 + c^2 y^2.$$
Because $b-c<a<b+c$, 
$$(a^2-b^2-c^2)yz \leq |a^2-b^2-c^2| |yz| < |2bc||yz| \leq b^2 z^2 + c^2 y^2.$$
The last inequality is AM-GM inequality. Hence the problem is proved. 
</s>