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设函数\(f\left(x\right)=\mathrm{e}^x-1-ax\).
\((1)\) 若\(x\geq0\),\(f\left(x\right)\geq0\),求\(a\)的取值范围;
\((2)\)若\(x>0\)且\(m\geq1\),证明：\(f\left(x\right)\geq\dfrac{x^2}{\ln\left(x+m\right)}-ax\).
(1) 因为\(f\left(x\right)=\mathrm{e}^x-1-ax\),则\(f{^{\prime}}\left(x\right)=\mathrm{e}^x-a\),
当\(a\le1\),则\(f{^{\prime}}\left(x\right)\geq0\),\(f\left(x\right)=\mathrm{e}^x-1-ax\)在\(\left[0,+\infty\right)\)单调递增,
而\(f\left(0\right)=0\),所以\(f\left(x\right)\geq0\)；
当\(a>1\),则当\(x\in\left(0,\ln a\right)\),\(f{^{\prime}}\left(x\right)=\mathrm{e}^x-a<0\),
即\(f\left(x\right)=\mathrm{e}^x-1-ax\)在\(x\in\left(0,\ln a\right)\)单调递减,\(f\left(x\right)<f\left(0\right)=0\),不符合题意,
综上可得\(a\)的取值范围为\(\left(-\infty,1\right]\).
(2).因为\(x>0\),\(m\geq1\),\(f\left(x\right)=\mathrm{e}^x-1-ax\),
所以要证明\(f\left(x\right)\geq\dfrac{x^2}{\ln\left(x+m\right)}-ax\),
只需
证明\(\dfrac{\mathrm{e}^x-1}{x}\geq\dfrac{x}{\ln\left(x+m\right)}\),
只需证\(\dfrac{\mathrm{e}^x-1}{x}\geq\dfrac{x}{\ln\left(x+1\right)}\),
即证\(\dfrac{\mathrm{e}^x-1}{x}\geq\dfrac{\mathrm{e}^{\ln\left(x+1\right)}-1}{\ln\left(x+1\right)}\),
设\(g\left(x\right)=\dfrac{\mathrm{e}^x-1}{x}\),\(x\geq0\),
所以
\(g^{\prime}\left(x\right)=\dfrac{\left(x-1\right)\mathrm{e}^x+1}{x^2}\),
令\(h\left(x\right)=\left(x-1\right)\mathrm{e}^x+1\),因为\(h^{\prime}\left(x\right)=x\mathrm{e}^x\geq0\),
所以\(h\left(x\right)=\left(x-1\right)\mathrm{e}^x+1\)在\(\left[0,+\infty\right)\)单调递增,
所以\(h\left(x\right)\geq h\left(0\right)=0\),所以\(g^{\prime}\left(x\right)\geq0\),
所以\(g\left(x\right)\)在\(\left[0,+\infty\right)\)单调递增,
令\(t\left(x\right)=x-\ln\left(x+1\right)\),\(x\geq0,t^{\prime}\left(x\right)=1-\dfrac{1}{x+1}=\dfrac{x}{x+1}\geq0\),
所以\(t\left(x\right)\)在\(\left[0,+\infty\right)\)上单调递增,
所以当\(x>0\)时,\(t\left(x\right)>t\left(0\right)=0\),
即\(x\geq \ln\left(x+1\right)\)成立,
所以\(g\left(x\right)\geq g\left(\ln\left(x+1\right)\right)\),
即\(\dfrac{\mathrm{e}^x-1}{x}\geq\dfrac{\mathrm{e}^{\ln\left(x+1\right)}-1}{\ln\left(x+1\right)}\)
所以\(f\left(x\right)\geq\dfrac{x^2}{\ln\left(x+m\right)}-ax\)成立.