难度很大的估值问题

已知\(f(x)=\dfrac{x^2}{2}+\cos x\)

(1)求\(f(x)\)最小值

(2)当\(0<x<1\)时，若\(\dfrac{\sin x}{x}>\dfrac{a}{x+2}\)恒成立，求\(a\)范围

(3)证明:\(\displaystyle\sum\limits_{k=1}^{n}\dfrac{\cos\dfrac{1}{4\sqrt{k}-1}}{2\sqrt{k}-1}>\sqrt{n+1}-1\)

解

(1)\(f(x)\geq \dfrac{x^2}{2}+1-\dfrac{x^2}{2}=1\)

(2) 因\(\dfrac{a}{x+2}<\dfrac{\sin x}{x}<1\)

从而\(a<x+2<2\)

则\(a\leq 2\)，\(\dfrac{\sin x}{x}>\dfrac{a}{x+2}\)恒成立

现说明\(a>2\)不合题

即\((x+2)\sin x-ax>0\)恒成立

记\(\varphi(x)=(x+2)\sin x-ax,\varphi(0)=0\)

\(\varphi^{\prime}(x)=\sin x+(x+2)\cos x-a,\varphi^{\prime}(0)=2-a\)

\(\varphi^{\prime\prime}(x)=2\cos x-(x+2)\sin x>2\left(1-\dfrac{x^2}{2}\right)-(x+2)x=2(1-x-x^2)\)
则得到\(\varphi^{\prime\prime}(x)\)在\(\left(0,\dfrac{1}{2}\right)\)上大于\(0\)

从而\(\varphi^{\prime}(x)\)在\(x\in \left(0,\dfrac{1}{2}\right)\)上单调递增

则\(\varphi^{\prime}(x)>\varphi^{\prime}(0)=2-a<0\)

从而，一定存在一个区间\((0,m)\)，使得\(\varphi^{\prime}(x)\)

在\((0,m)\)上为负

从而\(\varphi(x)<\varphi(0)=0\)，不合题

耍赖的写法

因\(\lim\limits_{x\to 0}\dfrac{\sin x}{x}=1\)

两边同取\(x\to 0\)有\(1>\dfrac{a}{2}>1\)，矛盾！

从而\(a\leq 2\)

(3) 由(2)取\(a=2\)有\(\sin x>\dfrac{2x}{x+2}\)，

则有\(x>\sin x>\dfrac{2x}{x+2}\)

即\(\dfrac{1}{4\sqrt{k}-1}>\sin\dfrac{1}{4\sqrt{k}-1}>\dfrac{\dfrac{2}{4\sqrt{k}-1}}{\dfrac{1}{4\sqrt{k}-1}+2}\)

\(\dfrac{\cos\dfrac{1}{4\sqrt{k}-1}}{2\sqrt{k}-1}>\dfrac{\cos\dfrac{1}{4\sqrt{k}-1}}{4\sqrt{k}-1}>\sin\dfrac{1}{4\sqrt{k}-1}\cos\dfrac{1}{4\sqrt{k}-1}=2\sin\dfrac{2}{4\sqrt{k}-1}\)

而\(2\sin\dfrac{2}{4\sqrt{k}-1}>2\cdot \dfrac{\dfrac{4}{4\sqrt{k}-1}}{\dfrac{2}{4\sqrt{k}-1}+2}=2\cdot\dfrac{1}{2\sqrt{k}}>2\cdot \dfrac{1}{\sqrt{k}+\sqrt{k+1}}=2(\sqrt{k+1}-\sqrt{k})\)

取\(k=1,2,3\cdots n\)有

\(\displaystyle\sum\limits_{k=1}^{n}\dfrac{\cos\dfrac{1}{4\sqrt{k}-1}}{2\sqrt{k}-1}>2(\sqrt{n+1}-1)>\sqrt{n+1}-1\)