Prove that (a - x)^{6} - 3a(a - x)^{5} + 5/2 a^{2}(a - x)^{4} - 1/2 a^{4}(a - x)^{2} < 0 for 0 < x < a.

**Solution**

Easy.

Change variables to t = 1 - x/a and the polynomial becomes a^{6}(t^{6} - 3t^{5} + 5/2 t^{4} - 1/2 t^{2}). This obviously has a factor t^{2}, and almost obviously (t - 1). Dividing these out, we see that the resulting cubic has another factor (t - 1). So we can write the original as a^{6}t^{2}(t - 1)^{2}(t(t - 1) - 1/2), which evidently has the same sign as t(t - 1) - 1/2. But that is clearly negative for t between 0 and 1.

© John Scholes

jscholes@kalva.demon.co.uk

15 Sep 1999