再提供一个证法： x，y为正数，x^+y^/2=1,则x(1+y^)^(1/2)的最大值是?
设x=cosα,y=√2sinα,(α为锐角)
则x(1+y^)^(1/2)=cosα[1+(√2sinα)^2]^(1/2)={2(cosα)^2·[1+2(sinα)^2]}^(1/2)/√2≤{2(cosα)^2+[1+2(sinα)^2]}/√2=3/√2
∴ x(1+y^)^(1/2)的最大值是3√2/2
x^+y^/2=1 ==> y^/2 = 1 -x^ >= 0 1 >= x^ x(1+y^)^(1/2) = x[1+2(1 -x^)]^(1/2) = [-2x^4 +3*x^2]^(1/2) = [-2(x^ -3/4)^ +9/8]^(1/2) <= (9/8)^(1/2) = 3*2^(1/2)/4 最大值 = 3*2^(1/2)/4
答：因x、y为正数,故x^2+y^2=1 ==> x=根号(1-y^2);因此,x*[根号(1+y^2)]=根号(1-y^2)(1+y^2)=<[(1-y^2)+(...详情>>
