## Thursday, December 1, 2005

### Factor That! Topic: Factoring. Level: AMC/AIME/Olympiad.

Problem #1: Find all values for which we have $x^3+3x^2+3x+1$ a positive number.

Solution Consider rewriting $x^3+3x^2+3x+1 = (x+1)^3$. Then we know $x+1$ has to be positive so $x>-1$. QED.

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Comment: This is a quite simple application of a very powerful technique, which we can derive some very cool inequalities from as well.

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Problem #2: Prove that $x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \le x^3+y^3+z^3+3xyz$ for positive reals $x,y,z$.

Solution: Notice that something like Rearrangement doesn't quite cut it here because we have the $3xyz$ on the LHS. So we consider factoring this into

$0 \le x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)$.

Well this looks pretty good, all symmetric and stuff. But how do we prove it? Let's try first assuming WLOG that $x \ge y \ge z$ because of symmetry. Then

$x(x-y)(x-z)+y(y-z)(y-x) \ge 0$

and

$z(z-x)(z-y) \ge 0$.

Add them up and we have our desired inequality. QED.

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Comment: This is known as Schur's Inequality and can be generalized to $x^n(x-y)(x-z)+y^n(y-z)(y-x)+z^n(z-x)(z-y) \ge 0$ for $n$ a positive integer with equality at $x=y=z$ or $x=0, y=z$ and permutations of that.

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Practice Problem #1: Prove that, given three positive reals $a,b,c$, we have $x^2y^2+x^2z^2+y^2z^2+y^2x^2+z^2x^2+z^2y^2 \le x^4+y^4+z^4+x^2yz+y^2zx+z^2xy$.

Practice Problem #2: Factor $a^3+b^3+c^3-3abc$.

1. The second problem is pretty well-known:

a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - (a*b + b*c + a*c)).

It came in handy on a USAMTS problem last year, actually, because the
second term is always greater than or equal to zero.

Sorry for the double-post; I was sloppy and made a mistake on my first one (you can delete it).

2. [...] which is just Schur’s Inequality, so the result is proved. QED. [...]