已知函数f(x)=sinx-xcosx+1/2,不等式f(x)

sinx-xcosx+1/20在(0,+∞)上恒成立令g(x)=1/3*x^3-sinx+xcosx+a-0。5则g'(x)=x^2-xsinx=x(x-sinx)令h(x)=x-sinx,则h'(x)=1-cosx在(0,+∞)上恒>0,所以h(x)在(0,+∞)上单调递增当x=0时x-sinx=0,所以当x>0时x-sinx>0所以g'(x)=x(x-sinx)在(0,+∞)上恒>0,所以g(x)在(0,+∞)上单调递增g(0)=a-0。5>=0所以a>=0。5