# Prove that $a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$

Suppose that $a,b,c$ are the lengths of the sides of a triangle. Prove that $$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$$

Attempt

It's hard to do much simplification here so we are going to have to use that $a > b+c, b>a+b, c>a+c.$ Using that fact we obtain that $$b^2(c-b)+c^2(b-c) > a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c).$$ Then should I use AM-GM to finish it or what inequality?

Since $a, b, c$ are the lengths of the sides of a triangle, we can let $a=x+y, b=y+z, c=z+x$, where $x, y, z>0$. This is known as Ravi Substitution.

Substituting, we have:

$2(x+y)^2z+2(y+z)^2x+2(z+x)^2y\leq3(x+y)(y+z)(z+x)$

Expanding, we have:

$$2\sum_{sym}x^2y+12xyz\leq6xyz+3\sum_{sym}x^2y$$

or

$$6xyz\leq\sum_{sym}x^2y$$

which is clearly true by AM-GM inequality.

• This substitution is called Ravi substitution. – user236182 Jan 2 '16 at 14:40

It simplifies to $$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b) \ge 0$$ which is Schur's Inequality for $r=1$.