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 $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$

We'll prove that your inequality is true for all reals $p$, $q$ and $r$ 
and $a$, $b$ and $c$ are lengths-sides of triangle.
Indeed,
$$a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)=$$
$$=\frac{1}{2}\left((p-q)^2(a^2+b^2-c^2)+(p-r)^2(a^2+c^2-b^2)+(q-r)^2(b^2+c^2-a^2)\right)\geq0$$
because $\sum\limits_{cyc}(a^2+b^2-c^2)=a^2+b^2+c^2>0$ and
$$\sum\limits_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2)=\sum\limits_{cyc}(2a^2b^2-a^4)=16S^2>0$$
Done!
