X1≠X2
f(x1)≠f(x2)
1/2（f（X1）+f（X2））-f（(X1+X2)/2）
=1/2(2^x1+2^x2)-2^[(x1+x2)/2]
=1/2{2^x1+x^x2-2*2^[(x1+x2)/2]}
=1/2[2^(x1/2)-2^(x2/2)]^2>0
1/2（f（X1）+f（X2））＞f（(X1+X2)/2）