已知函数\(f\left(x\right)=\left(x+1\right)\ln x-ax+a\left(a\in \mathbb{R}\right)\)
(1) 若\(a=2\),试判断\(f\left(x\right)\)的单调性,并证明你的结论
(2) 设\(0<a\le1\),求证：\(\mathrm{e}^{ax}-1>f\left(x\right)+a\left(3x-1\right)-\left(x+1\right)\ln x-\ln\left(x+1\right)\)
(1). 若\(a=2\),则\(f\left(x\right)=\left(x+1\right)\ln x-2x+2\left(x>0\right)\),\(f^\prime\left(x\right)=\ln x+\dfrac{x+1}{x}-2=\ln x+\dfrac{1}{x}-1\left(x>0\right)\),
令\(\varphi\left(x\right)=f^\prime\left(x\right)\),则\(\varphi\left(x\right)\)的定义域为\(\left(0,+\infty\right)\),\(\varphi^\prime\left(x\right)=\dfrac{1}{x}-\dfrac{1}{x^2}=\dfrac{x-1}{x^2}\),
令\(\varphi^\prime\left(x\right)>0\),解得\(x>1\)；令\(\varphi^\prime\left(x\right)<0\),解得\(0<x<1\)；
则\(\varphi\left(x\right)\)在\(\left(0,1\right)\)上单调递减,在\(\left(1,+\infty\right)\)上单调递增,可得\(\varphi\left(x\right)\geq\varphi\left(1\right)=0\),
即\(f^\prime\left(x\right)\geq0\)对\(\forall x\in\left(0,+\infty\right)\)恒成立,
故\(f\left(x\right)\)在\(\left(0,+\infty\right)\)上单调递增.
(2).由题意可得：\(f\left(x\right)+a\left(3x-1\right)-\left(x+1\right)\ln x-\ln\left(x+1\right)=2ax-\ln\left(x+1\right)\),
则\(\mathrm{e}^{ax}-1>f\left(x\right)+a\left(3x-1\right)-\left(x+1\right)\ln x-\ln\left(x+1\right)\),即\(\mathrm{e}^{ax}-1>2ax-\ln\left(x+1\right)\),
可得\(\mathrm{e}^{ax}+\ln\left(x+1\right)-2ax-1>0\),
故原题意等价于\(\mathrm{e}^{ax}+\ln\left(x+1\right)-2ax-1>0\left(x>0\right)\),
令\(F\left(x\right){\mathrm{=e}}^{ax}-2ax+\ln\left(x+1\right)-1\),则\(F^\prime\left(x\right)\mathrm{=}a\mathrm{e}^{ax}+\dfrac{1}{x+1}-2a\),
发现\(F(0)=0\),若能证明\(F^{\prime}\geq 0\),那么结论便得证(这只是一个必要条件).
由结论\(e^x\geq x+1\)有
因为\(0<a\le1,x>0\)，则\(\mathrm{e}^{ax}>ax+1\geq ax+a\)
可得\(F^\prime\left(x\right)=a\mathrm{e}^{ax}+\dfrac{1}{x+1}-2a>a^2\left[\left(x+1\right)+\dfrac{\dfrac{1}{a^2}}{x+1}\right]-2a \)
因为当\(t>0,a>0\)时，则\(t+\dfrac{\dfrac{1}{a^2}}{t}\geq2\sqrt{t\times\dfrac{\frac{1}{a^2}}{t}}=\dfrac{2}{a}\)，当且仅当\(t=\dfrac{\dfrac{1}{a^2}}{t}\)，即\(t=\dfrac{1}{a}\)时，等号成立
即对\(\forall t>0,a>0\)，均有\(t+\dfrac{\dfrac{1}{a^2}}{t}\geq\dfrac{2}{a}\)，
故当\(x>0,0<a\le1\)，即\(x+1>1\)，可得\(\left(x+1\right)+\dfrac{\dfrac{1}{a^2}}{x+1}\geq\dfrac{2}{a}\)，
故\(F^\prime\left(x\right)>a^2\left[\left(x+1\right)+\dfrac{\dfrac{1}{a^2}}{x+1}\right]-2a\geq a^2\times\dfrac{2}{a}-2a=0\)，
则\(F\left(x\right)\)在\(\left(0,+\infty\right)\)上单调递增，可得\(F\left(x\right)>F\left(0\right)=0\).
故\(\mathrm{e}^{ax}-2ax+\ln\left(x+1\right)-1>0\)，即证.