设函数 f(x)=excosx,g(x)f\left( x\right) = {\mathrm{e}}^{x}\cos x,g\left( x\right)f(x)=excosx,g(x) 为 f(x)f\left( x\right)f(x) 的导函数.(I) 求 f(x)f\left( x\right)f(x) 的单调区间;(II) 当 x∈[π4,π2]x \in \left\lbrack {\frac{\pi }{4},\frac{\pi }{2}}\right\rbrackx∈[4π,2π] 时,证明 f(x)+g(x)(π2−x)≥0f\left( x\right) + g\left( x\right) \left( {\frac{\pi }{2} - x}\right) \geq 0f(x)+g(x)(2π−x)≥0 ;(III) 设 xn{x}_{n}xn 为函数 u(x)=f(x)−1u\left( x\right) = f\left( x\right) - 1u(x)=f(x)−1 在区间 (2nπ+π4,2nπ+π2)\left( {{2n\pi } + \frac{\pi }{4},{2n\pi } + \frac{\pi }{2}}\right)(2nπ+4π,2nπ+2π) 内的零点,其中 n∈Nn \in Nn∈N ,证明 2nπ+π2−xn<e−2nπsinx0−cosx0{2n\pi } + \frac{\pi }{2} - {x}_{n} < \frac{{e}^{-{2n\pi }}}{\sin {x}_{0} - \cos {x}_{0}}2nπ+2π−xn<sinx0−cosx0e−2nπ .