已知f(x)=cos2x,当0<a<b<1,问f[(a+b)/2]>[f(a)+f(b)]/2是否成立

f[(a+b)/2]-[f(a)+f(b)]/2]=cos(a+b)-[(cos2a+cos2b)/2]=cos(a+b)-cos(a+b)cos(a-b)=cos(a+b)[1-cos(a-b)]…(*)
(1) 0<a<b≤π/4时, 0<a+b<π/2, -π/4<a-b<π/4, cos(a+b)>0, cos(a-b)<1, ∴ (*)式>0, f[(a+b)/2]>[f(a)+f(b)]/2
(2) π/4≤a<b<1<π/2, π/2<a+b<π, -π/4<a-b<π/4, cos(a+b)<0, cos(a-b)<1, ∴ (*)式<0, f[(a+b)/2]<[f(a)+f(b)]/2