求函数y=根号下x一1加根号下9一3x的最大值

y=√(x-1)+√(9-3x)
=√(x-1)+√3*√(3-x)
≤√[(1+3)*(x-1)+3-x)]
=2√2
解2：令m=√(x-1),n=√(9-3x)
则m^2=x-1,n^2/3=3-x
相加得m^2+n^2/3=2,m^2/2+n^2/6=1
m=√2cosθ,n=√6sinθ
则y=m+n=2√2sin(θ+α)≤2√2