已知函数f(x)=a(e x +a)−xf(x)=a(e^x+a)-xf(x)=a(ex+a)−x （1）讨论f(x)f(x)f(x)的单调性 （2）证明：当a>0a>0a>0时，求证：f(x)>2ln⁡a+ 3 2f(x)>2\ln a+\dfrac 32f(x)>2lna+23​

解：\quad(1)f ′ (x)=ae x −1f'(x)=ae^x-1f′(x)=aex−1

\qquad①a>0a>0a>0时，x=−ln⁡ax=-\ln ax=−lna时f ′ (x)=0f'(x)=0f′(x)=0

f(x)\qquad f(x)f(x)在[−ln⁡a,+∞)[-\ln a,+\infty)[−lna,+∞)上单调递增，在(−∞,−ln⁡a](-\infty,-\ln a](−∞,−lna]上单调递减

\qquad②a≤0a\leq 0a≤0时，f(x)f(x)f(x)在(−∞,+∞)(-\infty,+\infty)(−∞,+∞)上单调递减

\quad(2)由(1)得x=−ln⁡ax=-\ln ax=−lna时f(x)f(x)f(x)取最小值

\qquad题目即证f(−ln⁡a)>2ln⁡a+ 3 2f(-\ln a)>2\ln a+\dfrac 32f(−lna)>2lna+23​

\qquad即1+a 2 +ln⁡a>2ln⁡a+ 3 21+a^2+\ln a>2\ln a+\dfrac 321+a2+lna>2lna+23​，a 2 −ln⁡a− 1 2>0a^2-\ln a-\dfrac 12>0a2−lna−21​>0

\qquad令g(a)=a 2 −ln⁡a− 1 2g(a)=a^2-\ln a-\dfrac 12g(a)=a2−lna−21​，则g ′ (a)=2a− 1 ag'(a)=2a-\dfrac 1ag′(a)=2a−a1​

\qquad当a=2 2a=\dfrac{\sqrt 2}{2}a=22​​时g ′ (a)=0g'(a)=0g′(a)=0

g(a)\qquad g(a)g(a)在[2 2,+∞)[\dfrac{\sqrt 2}{2},+\infty)[22​​,+∞)上单调递增，在(−∞,2 2](-\infty,\dfrac{\sqrt 2}{2}](−∞,22​​]上单调递减

\qquad所以g(a)≥g(2 2)=−ln⁡2 2=ln⁡2 >0g(a)\geq g(\dfrac{\sqrt 2}{2})=-\ln\dfrac{\sqrt 2}{2}=\ln\sqrt 2>0g(a)≥g(22​​)=−ln22​​=ln2​>0

\qquad即a 2 −ln⁡a− 1 2>0a^2-\ln a-\dfrac 12>0a2−lna−21​>0

\qquad得证f(x)>2ln⁡a+ 3 2f(x)>2\ln a+\dfrac 32f(x)>2lna+23​