x 1 & lt； X2 > H(x2) with h (x 1) indicates that the function is monotonically decreasing.

Let x 1=x2=0 to get h(0)=0.

And h (1/4) = h (1/2)+h (1/2) = 2.

h(-x)+h(3-x)& lt； -2=-h( 1/4)

h[(-x)(3-x)]+h( 1/4)& lt； =0

h[(x^2-3x)/4]<； =h(0)

That is, (x2-3x)/4 >; =0

x(x-3)>=0

The direct solution is: x & gt=3 or x.