The cubic x3 + ax2 + bx + c has real coefficients and three real roots r ≥ s ≥ t. Show that k = a2 - 3b ≥ 0 and that √k ≤ r - t.
Solution
a2 - 3b = (r + s + t)2 - 3(rs + st + tr) = r2 + s2 + t2 - (rs + st + tr). By Cauchy-Schwartz we have (rs + st + tr)2 ≤ (r2 + s2 + t2)2, so r2 + s2 + t2 ≥ |rs + st + tr| ≥ (rs + st + tr). Hence a2 - 3b ≥ 0.
a2 - 3b ≤ (r - t)2 is the same as r2 + s2 + t2 - rs - st - tr ≤ r2 - 2rt + ts or s2 + rt - rs - st ≤ 0 or (r - s)(s - t) ≥ 0, which is true since r ≥ s and s ≥ t. So a2 - 3b ≤ (r - t)2. Taking the non-negative square roots, we get the required result.
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002