If $a$, $b$ and $c$ are sides of a triangle, then prove that $a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c)$ $\leqslant$ $3abc$

Question

Let $a$, $b$ and $c$ be the sides of a triangle. Prove that $$a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c) \leqslant 3abc$$

SOURCE: BANGLADESH MATH OLYMPIAD (Preparatory Question.)

I am very new and novice at this problem. I did a little try but couldn't succeed because I was unable to substitute the left term of the inequality into formula. I know a formula that

$a^\text{3} + b^\text{3} + c^\text{3} - 3abc$ = $(a+b+c)(a^\text{2} + b^\text{2} + c^\text{2} - ab -bc - ca)$. But how to use this formula in that case isn't known to me. And how to show the relation of the both side and when they will become equal?

A small help will be enough for me. Thanks in advance.

Answer 1

Using the Ravi substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that $$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz\geq 0$$ But this is AM-GM:

$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2\geq 6\sqrt[6]{x^6y^6z^6}=6xyz$$

Answer 2

It's true for all non-negatives $a$, $b$ and $c$.

Indeed, since our inequality is symmetric, we can assume that $a\geq b\geq c$ and we obtain: $$3abc-\sum_{cyc}a^2(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0.$$