## Wednesday, December 23, 2015

### Prove $\small a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$ for all real $a,\,b,\,c\ge 0$: Second Attempt

For reals $a,\,b,\,c\ge 0$, prove the inequality

$a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$

and state when the equality holds.

From our previous attempt, we know AM-GM inequality wouldn't help, so we try to expand both sides of the equation and see where that leads us:

$a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$

$a(a^2-2ac+c^2)+b(b^2-2bc+c^2)\ge (ab-ac-bc+c^2)(a+b-c)$

$a^3-2a^2c+ac^2+b^3-2b^2c+bc^2$

$\ge a^2b+ab^2-abc-a^2c-abc+ac^2-abc-b^2c+bc^2+ac^2+bc^2-c^3$

Rearrange the terms we get:

[MATH]\color{yellow}\bbox[5px,green]{a^3+b^3+c^3+3abc\ge a^2(b+c)+b^2(a+c)+c^2(a+b)}[/MATH]

I hope you've developed a good sense of familiarity with Schur's inequality that says:

For non-negative real numbers $x,$ $y,$ $z$ and a positive number $t,$

$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\ge 0.$

The equality holds in two cases:

1. $x=y=z,$ or

2. One fo them is $0,$ while the other two are equal.

For $t=1$, Schur's inequality can be rearranged into

[MATH]\color{yellow}\bbox[5px,purple]{x^3+y^3+z^3+3xyz\ge x^2(y+z)+y^2(x+z)+z^2(x+y)}[/MATH]

Therefore
[MATH]\color{yellow}\bbox[5px,green]{a^3+b^3+c^3+3abc\ge a^2(b+c)+b^2(a+c)+c^2(a+b)}[/MATH] is actually the Schur's inequality when $t=1$. Equality occurs when $a=b=c$ or when two of $a,\,b$ or $c$ are equal, and the third is a zero.