已知f(x)=ax^2+4x-2,若对任意x1,x2属于实数且x1x2不相等都有f((x1+x2)/2)<(f(x1)+f(x2))/2.

若a=0
则f(x)=4x-2
f((x1+x2)/2)<(f(x1)+f(x2))/2.
2(x1+x2)-2<(4x1-2+4x2-2)/2
2x1+2x2-2<2x1+2x2-2
不成立

a不等于0
f((x1+x2)/2)<(f(x1)+f(x2))/2.
a[(x1+x2)/2]^2+4(x1+x2)/2-2<(ax1^2+4x1-2+ax2^2+4x2-2)/2
a(x1+x2)^2/4+2(x1+x2)-2<a(x1^2+x2^2)/2+2(x1+x2)-2
所以a(x1+x2)^2<2a(x1^2+x2^2)

因为x1不等于x2
所以(x1-x2)^2>0
所以x1^2-2x1x2+x2^2>0
x1^2+x2^2>2x1x2
(x1^2+x2^2)+x1^2+x2^2>(x1^2+x2^2)+2x1x2
所以(x1+x2)^2<2(x1^2+x2^2)
要满足a(x1+x2)^2<2a(x1^2+x2^2)
所以a>0