每日导数45

估值问题要放缩

设函数\(f(x)=\sin x-x\cos x,g(x)=\left(1+\dfrac{x^2}{2}\right)\cos x\)

(1) 当\(x\in[0,\pi]\)时，证明:\(f(x)\geq 0\)

(2) 当\(x\in[-\pi,\pi]\)时，求\(g(x)\)的值域

(3)若数列\(\{a_n\}\)满足\(a_1=1,a_{n+1}=a_n\cos a_n,a_n>0\)，证明:

\((3a_1+a_2+a_3+\cdots+a_n)\cos a_1\cos a_2\cos a_3\cdots\cos a_n<2\)
解

(1)\(f^{\prime}(x)=x\sin x\)，从而\(f(x)\geq f(0)=0\)

(2)因\(g(x)\)是偶函数，考虑\(x\in[0,\pi]\)

\(g^{\prime}(x)=-\sin x+x\cos x-\dfrac{x^2}{2}\sin x\)

由\((1)\),\(x\cos x-\sin x\leq 0\)

则\(g^{\prime}(x)\leq 0\)，从而\(g(x)\)单调递减

则\(x\in[0,\pi]\)，\(g(x)\in\left[-1-\dfrac{\pi^2}{2},1\right]\)

则\(g(x)\)值域为\(\left[-1-\dfrac{\pi^2}{2},1\right]\)

(3) 因\(a_{n+1}=a_n\cos a_n\)则\(\cos a_n=\dfrac{a_{n+1}}{a_n}\)

从而\(\cos a_1\cos a_2\cos a_3\cdots\cos a_n=\dfrac{a_{n+1}}{a_1}=a_{n+1}\)

则原不等式变为

\[(3a_1+a_2+a_3+\cdots+a_n)<\dfrac{2}{a_{n+1}} \]

由(1)\(x\cos x\leq \sin x\)

则\(a_{n+1}=a_n\cos a_n\leq \sin a_n\leq a_n\)

从而\(a_n\)单调递减

因\(a_1=1\)，则\(0<a_n<1\)

由(2)得\(g(x)\)是单调递减的

从而\(\left(1+\dfrac{x^2}{2}\right)\cos x<1,x\in(0,1)\)

则\(\cos x<\dfrac{1}{1+\dfrac{x^2}{2}}\)

则\(a_{n+1}<\dfrac{a_n}{1+\dfrac{a_n^2}{2}}\)

即\(a_{n+1}<\dfrac{2a_n}{2+a_n^2}\)

两边取倒数有

\(\dfrac{1}{a_{n+1}}>\dfrac{1}{a_n}+\dfrac{a_n}{2}\)

从而\(a_n<2\left(\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n}\right)\)

则
\(3a_1+a_2+a_3+\cdots+a_n<2a_1+2\left(\dfrac{1}{a_2}-\dfrac{1}{a_1}+\dfrac{1}{a_3}-\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n}\right)=2a_1+\dfrac{2}{a_{n+1}}-\dfrac{2}{a_1}=\dfrac{2}{a_{n+1}}\)
因\(a_1=1\)，则原不等式得证.