prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$

Question

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

Answer 1

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!

We used the following lemma.

Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that $x+y+z\geq0$ and $xy+xz+yz\geq0.$ Prove that: $$x(b-c)^2+y(c-a)^2+z(a-b)^2\geq0.$$

Indeed, since $x+y+z\geq0$ so $\sum\limits_{cyc}(x+y)\geq0$ and we can assume that $x+y\geq0$(if $x+y<0$,$x+z<0$ and $y+z<0$ so we obtain a contradiction).

Thus, we need to prove that $$x(b-c)^2+y(b-c+a-b)^2+z(a-b)^2\geq0$$ or $$(x+y)(b-c)^2+2y(b-c)(a-b)+(y+z)(a-b)^2\geq0.$$ Now, if $x+y=0$, so from $x+y+z\geq0$ we obtain $z\geq0$ and from $xy+xz+yz\geq0$ we obtain $xy+z(x+y)\geq0$ or $xy\geq0$, which gives $-x^2\geq0$ and $x=y=0$ and our inequality is true in this case.

Let $x+y>0$.

Thus, it's enough to prove that $$y^2-(x+y)(y+z)\leq0$$ or $$xy+xz+yz\geq0,$$ which ends the proof.

Answer 2

Note that substitution $p'=p-t, q'=q-t, r'=r-t$, where $t$ is any real number leads to an equivalent form $$a^2(p'-q')(p'-r')+b^2(q'-p')(q'-r')+c^2(r'-p')(r'-q')\ge 0.$$ This, as you proved already, is equivalent to $$(a^2 p'^2+b^2 q'^2 + c^2 r'^2) \ge p'r'(a^2+c^2-b^2)+q'r'(b^2+c^2-a^2)+p'q'(a^2+b^2-c^2).$$ However, since the inequality is symmetric with respect to $p,q,r$, we can assume that $r = \min\{p,q,r\}$. Putting $t=r$ we get $r'=0$ and so the inequality reduces to $$a^2 p'^2+b^2 q'^2 \ge p'q'(a^2+b^2-c^2),$$ where $p', q'$ are nonnegative (because $p,q\ge r$).

Triangle inequality yieds $c>|a-b|$, so $c^2 > a^2-2ab+b^2$ and therefore $$a^2+b^2-c^2 < 2ab.$$

Since $p',q'$ are nonnegative, we have that $$p'q'(a^2+b^2-c^2) \le 2p'q'ab.$$

Thus it suffices to prove that $$a^2p'^2+b^2q'^2 \ge 2p'q'ab.$$

But this is obvious: $$a^2p'^2 + b^2q'^2 - 2p'q'ab = (ap'-bq')^2 \ge 0.$$

Answer 3

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.