9．已知函数f(x)=x2+πcosxf(x)=x^{2}+\pi \cos xf(x)=x2+πcosx．（1）求函数f(x)f(x)f(x)的最小值；（2）若函数g(x)=f(x)−ag(x)=f(x)-ag(x)=f(x)−a在(0,+∞)(0,+\infty )(0,+∞)上有两个零点x1x_{1}x1，x2x_{2}x2，且x1<x2x_{1}<x_{2}x1<x2，求证：x1+x2<πx_{1}+x_{2}<\pix1+x2<π．