第 1 題
f(x) = x^3 + ax^2 + bx + 1
由於 a ≧ 0,b ≧ 0
易知 x ≧ 0,f(x) ≧ 1,故 f(x) = 0 之三根均為負
令 f(x) = 0 之三根為 -p、-q、-r (p、q、r > 0)
pqr = 1
a = p + q + r ≧ 3(pqr)^(1/3) = 3
b = pq + qr + rp ≧ 3(pqr)^(2/3) = 3
等號均成立於 p = q = r
f(2) = 4a + 2b + 9 ≧ 12 + 6 + 9 = 27
[ 本帖最後由 thepiano 於 2014-4-17 07:25 PM 編輯 ]