(√(ma+nb))^2-(m√a+n√b)^2
=ma+nb-m^2 a-n^2 b-2mn√ab
=ma(1-m)+nb(1-n)-2mn√ab
=mna+mnb-2mn√ab
=mn(a+b-2√ab)
=mn(√a-√b)^2≥0
所以 (√(ma+nb))^2≥（m√a+n√b)^2
所以√(ma+nb)≥m√a+n√b