有界性，第三问有点问题，没说明数列的唯一性

已知函数\(f(x)=\sqrt{\dfrac{2x}{x+1}},g(x)=\dfrac{\sin x}{x}\)

(1)求\(f(x)\)单调增区间

(2)证明：\(-\dfrac{1}{4}<g(x)<1\)

(3)设\(x_1=\sqrt2,x_{n+1}=f(x_n)\)，证明：\(x_1x_2\cdots x_n<\dfrac{\pi}{2}\)

解

(1)定义域：\(2x(x+1)\geq 0\)得\(x<-1\)或\(x\geq 0\)

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

则\(f(x)\)的单调增区间为\(x<-1\)或\(x\geq 0\)

(2) \(g(x)\)的定义域\(x\neq 0\)

因\(\sin x<x\)，则\(g(x)<1\)

因\(g(x)\)是偶，考虑\(x>0\)，则左边即证:\(h(x)=\sin x+\dfrac{x}{4}>0\)

\(h^{\prime}(x)=\cos x+\dfrac{1}{4}\)，而\(\cos x\)在\(\left(0,\pi\right]\)上减，

利用\(\sin x\)的有界性，\(|\sin x|\leq 1\)，考虑\(\dfrac{x}{4}>1\)，\(x>4\)

则\(x>\dfrac{4\pi}{3}>4\)时，则\(f(x)>0\)

当\(x\in(0,\pi)\)时，\(h(x)>0\)

当\(x\in\left[\pi,\dfrac{4\pi}{3}\right]\)时，\(h(x)\)减，\(h(x)\geq h\left(\dfrac{4\pi}{3}\right)=\sin\dfrac{4\pi}{3}+\dfrac{\pi}{3}=\dfrac{\pi}{3}-\dfrac{\sqrt3}{2}>0\)

综上，\(\sin x+\dfrac{x}{4}>0\)得证

综上\(-\dfrac{1}{4}<g(x)<1\)

(3)\(x_{n+1}=\sqrt{\dfrac{2x_n}{x_n+1}}\)，两边平方\(x^2_{n+1}=\dfrac{2x_n}{x_n+1}\)

即\(\dfrac{1}{x^2_{n+1}}=\dfrac{x_n+1}{2x_n}=\dfrac{1}{2}+\dfrac{1}{2x_n}\)

即\(\dfrac{2}{x^2_{n+1}}=\dfrac{1}{x_n}+1\)

因\(x_1=\sqrt{2}\)，则\(a_1=\dfrac{1}{x_1}=\dfrac{\sqrt2}{2}=\cos\dfrac{\pi}{4}\)

而\(2\cos^2\dfrac{\pi}{2^{n+2}}=\cos^2\dfrac{\pi}{2^{n+1}}+1\)

从而\(\dfrac{1}{x_n}=\cos^2\dfrac{\pi}{2^{n+1}}\)

则\(x_n=\dfrac{1}{\cos\dfrac{\pi}{2^{n+1}}}\)

则\(x_1x_2x_3\cdots x_n=\dfrac{1}{\cos\dfrac{\pi}{2}\cos\dfrac{\pi}{2^{2}}\cos^2\dfrac{\pi}{2^{3}}\cdots\cos\dfrac{\pi}{2^{n+1}}}=\dfrac{2^n\sin\dfrac{\pi}{2^{n+1}}}{2^n\cos\dfrac{\pi}{2}\cos\dfrac{\pi}{2^{2}}\cos\dfrac{\pi}{2^{3}}\cdots\cos\dfrac{\pi}{2^{n+1}}}=\dfrac{2^n\sin\dfrac{\pi}{2^{n+1}}}{\sin\dfrac{\pi}{2}}=2^n\sin\dfrac{\pi}{2^{n+1}}\)
而\(2^n\sin\dfrac{\pi}{2^{n+1}}<2^n\dfrac{\pi}{2^{n+1}}=\dfrac{\pi}{2}\)

得证.