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.


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


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 $


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

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


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.


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. [...]