显然,x≥ 0时,2^x≥1,即g(x)≥ 0,
因为x1,x2≥ 0,所以2^x1-1≥ 0,2^x2-1≥ 0,
(2^x1-1)(2^x2-1)≥ 0,
2^x1*2^x2-2^x1-2^x2+1≥ 0,
2^(x1+x2)-1≥2^x1-2^x2-2,
即2^(x1+x2)-1≥ (2^x1-1)+(2^x2-1),
亦即g(x1+x2)≥ g(x1)+g(x2).