(√3b)^2=(1-a)(1+a)
3b^2=1-a^2
a^2+3b^2=1
令 a=sint,b=cost/√3
则 a+3b=sin t +√3cost
=2(1/2*sint +√3cost）
=2(sint*cosπ/3 +cost*sinπ/3）
=2sin(t+π/3)
-1<=sin(t+π/3)<=1
所以a+3b的最大值为:2