Question

f(x)=[ax/(x^2+1)]+a g(x)=alnx-x(a≠0) 求证：当a>0时，对于任意x1 x2∈(0,e]总有g(x1)<f(x2)成立

求解

Answer 1

当a>0时，对于任意x1, x2∈(0,e]总有g(x1)<f(x2)成立, 只须在(0, e]内, g(x)最大值小于f(x)的最小值
f(x) = ax/(x² + 1) + a

f'(x) = [a(x² + 1) - ax*2x]/(x² + 1)² = a(1 - x²)/(x² + 1)² = 0
x = 1 (不必考虑x = -1 < 0)
0 < x < 1: a > 0, 1 - x² > 0, (x² + 1)² > 0, f'(x) > 0
x > 1: a > 0, 1 - x² < 0, (x² + 1)² > 0, f'(x) < 0
f(1)为最大值
f(0) = a, f(e) = ae/(1 + e²) + a > a
f(0)在[0, e]内取最小值a

g(x) = alnx - x
g'(x) = a/x - 1 = 0
x = a
0 < x < a: a/x - 1 = (a - x)/x > 0, g'(x) > 0

x > a: a/x - 1 = (a - x)/x < 0, g'(x) < 0
g(a)为最大值
(i) a > e
g(e)在(0, e]内取最大值, g(e) = a - e < a (f(x)在[0, e]内的最小值)

(ii) 0 < a < e
lna < lne = 1

g(a)在(0, e]内取最大值, g(a) = alna - a = a(lna - 1) < a(1 - 1) = 0 < a (f(x)在[0, e]内的最小值)
证毕