題目有誤,是證明 (x-1)f(x) ≥ 0
定義域 x>0
f'(x) = lnx + (x+1)/x - 1 = lnx + 1/x
f''(x) = 1/x - 1/x² = (x-1)/x²
f''(x) = 0 得 x=1
∴f'(x) 在(0,1) 遞減,在(1,+∞)遞增
f'(x) ≥ f'(1) = 1 > 0
∴f(x) 單調(diào)遞增
當(dāng)x≥1時,x-1≥0
f(x)≥f(1) = 0
∴(x-1)f(x) ≥ 0
當(dāng) 0