If $a, b, c$ are the sidelengths of a triangle, show that $a^2b(a−b) +b^2c(b−c)+c^2a(c−a)\ge0$.

Question

After many days of work and some help with some helpful Math Stack Exchange community members, I have only one inequality homework question which remains unsolved:

If $a, b, c$ are the sidelengths of a triangle, show that $a^2b(a−b) +b^2c(b−c)+c^2a(c−a)\ge0$.

My attempt:

Let $a=y+z, b=z+x, c=x+y$. Then $x,y,z\ge0$.

But after substitute into the inequality, and expand, I still cannot use Muirhead.

This question is one of the starred question and I can't do it.

Can someone help me? Any help is appreciated!

Answer 1

Hint

Schur inequality might be helpful $$\sum_{cyc}a^2(a-b)(a-c) \geq 0$$

Now, the triangle inequalities like $a-c \leq b \Leftrightarrow a \leq b+c$ might be helpful.

Answer 2

Using the Ravi-substitution we get $$xy^3+x^3z+yz^3\geq xyz(x+y+z)$$ or $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z$$ Now using Cauchy Schwarz in Engelform: $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq \frac{(x+y+z)^2}{x+y+z}=x+y+z.$$

Answer 3

Let $c=\min\{a,b,c\}$.

Thus, we need to prove that $$\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\geq\frac{a^2b^2+a^2c^2+b^2c^2}{abc}$$ or $$\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{a^2}{c}-\frac{a^2}{b}+\frac{c^2}{b}-c\geq\frac{a^2b^2+a^2c^2+b^2c^2-abc(a+b+c)}{abc}$$ or $$\frac{(a-b)^2(a+b)}{ab}+\frac{(c^2-a^2)(c-b)}{bc}\geq\frac{(ac-bc)^2+(bc-ab)(ac-ab)}{abc}$$ or $$\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c-b)(a+c)}{bc}\geq\frac{(a-b)^2c}{ab}+\frac{(c-a)(c-b)}{c}.$$ Can you end it now?

Answer 4

The same inequality holds for bigger exponents i.e: $$\sum_{\text{cyc}}a^pb(a-b)\geq 0$$ for $p\geq 2$ and $a,b,c$ sides of a triangle. You can prove it easily by thinking of the expression as $f(p)$, a function of $p$, and take a derivative to show $f(p)$ is non-decreasing and so $$f(p)\geq f(2)\geq 0.$$