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按题意，b > 1, 且x² - ax + b = (x - 1)(x - b) = x² - (b+1)x + b
a = b + 1
ab = b(b + 1) = (b + 1/2)² - 1/4 > (1 + 1/2)² - 1/4 = 2
f(b) = 4/a - 2/b = 4/(b + 1) - 2/b = 2(b - 1)/[b(b + 1)]
f'(b) = [2b(b+1) - 2(b - 1)(2b + 1)]/[b(b + 1)]² = (-2b² + 4b + 2)/[b(b + 1)]²
f'(b) = 0, b² - 2b - 1 = 0
b = 1±√2, 不必考虑b = 1-√2 < 1
b = 1 +√2时, f(b)取极值6-4√2
b = 1时, f(1) = 0,
b -> +∞时, f(b) ->0
即0 < 4/a - 2/b ≤ 6-4√2
a = b + 1
ab = b(b + 1) = (b + 1/2)² - 1/4 > (1 + 1/2)² - 1/4 = 2
f(b) = 4/a - 2/b = 4/(b + 1) - 2/b = 2(b - 1)/[b(b + 1)]
f'(b) = [2b(b+1) - 2(b - 1)(2b + 1)]/[b(b + 1)]² = (-2b² + 4b + 2)/[b(b + 1)]²
f'(b) = 0, b² - 2b - 1 = 0
b = 1±√2, 不必考虑b = 1-√2 < 1
b = 1 +√2时, f(b)取极值6-4√2
b = 1时, f(1) = 0,
b -> +∞时, f(b) ->0
即0 < 4/a - 2/b ≤ 6-4√2
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谢谢啦
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