Prove that (a - x)6 - 3a(a - x)5 + 5/2 a2(a - x)4 - 1/2 a4(a - x)2 < 0 for 0 < x < a.
Solution
Easy.
Change variables to t = 1 - x/a and the polynomial becomes a6(t6 - 3t5 + 5/2 t4 - 1/2 t2). This obviously has a factor t2, 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 a6t2(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