端点效应与适当放缩

已知函数\(f(x)=e^x+\cos x-2,g(x)=\sin x\)

(1)证：当\(x>0\)时，\(g(x)<x<f(x)\)

(2)若\(x>0,f(x)+g(x)>ax\)恒成立，求\(a\)的取值范围.

解
(1)左边经典不等式，略

右边：\(x<e^x+\cos x-2\)

即证\(e^x+\cos x-2-x>0\)

记\(\varphi(x)=e^x+\cos x-2-x,\varphi^{\prime}(x)=e^x-\sin x-1>x+1-x-1=0\)

从而\(\varphi(x)>\varphi(0)=0\)

得证

（2)\(f(x)+g(x)=e^x+\cos x-2+\sin x\)

则\(e^x+\cos x-2+\sin x>ax\)恒成立

记\(\gamma(x)=e^x+\cos x-2+\sin x-ax\)

\(\gamma^{\prime}(x)=e^x-\sin x+\cos x-a,\gamma^{\prime\prime}(x)=e^x-\cos x-\sin x,\gamma^{\prime\prime\prime}(x)=e^x+\sin x-\cos x\)

\(\gamma(0)=0,\gamma^{\prime}(0)=2-a,\gamma^{\prime\prime}(0)=0\)

Case1当\(a\leq2\)，因\(\sin x\leq x,\cos x\geq 1-\dfrac{x^2}{2},e^x\geq 1+x+\dfrac{x^2}{2}\)

从而放缩得：\(\gamma^{\prime}(x)>1+x+\dfrac{x^2}{2}-x+1-\dfrac{x^2}{2}-a=2-a\geq 0\)

从而\(\gamma(x)\)单调递增，则\(\gamma(x)>\gamma(0)=0\)合题

Case2当\(a>2\)时，\(\gamma^{\prime}(0)=2-a<0\)

\(\gamma^{\prime\prime\prime}(x)=e^x+\sqrt2\sin\left(x-\dfrac{\pi}{4}\right)\)，在\(x\in\left(0,\dfrac{3\pi}{4}\right)\)上递增

从而\(\gamma^{\prime\prime\prime}(x)>\gamma^{\prime\prime\prime}(0)=0\)

则\(\gamma^{\prime\prime}(x)\)在\(x\in\left(0,\dfrac{3\pi}{4}\right)\)上单调递增，则\(\gamma^{\prime\prime}(x)>\gamma^{\prime\prime}(0)=0\)

则\(\gamma^{\prime}(x)\)在\(x\in\left(0,\dfrac{3\pi}{4}\right)\)上单调递增

因\(\gamma^{\prime}(0)<0\)，从而一定存在一个区间\((0,m)\)

使得在\(x\in(0,m)\)上，\(\gamma^{\prime}(x)<0\)

从而\(\gamma(x)\)在\(x\in(0,m)\)上单调递减

则\(\gamma(x)<\gamma(0)=0\)，矛盾！不合题

综上\(a\leq 2\)