a=f[(x1+x2)/2]=√[(x1+x2)/2]>0,--> a^2=(x1+x2)/2
b=[f(x1)+f(x2)]/2=(√x1+√x2)/2>0-->b^2=[x1+x2+2√(x1x2)]/4b
因此：f[（x1+x2）/2]>[f（x1）+f（x2）]/2