x>0y>0x+y=1,x^2+y^2+根号xy的最大值答案是9/8

Question

x>0y>0x+y=1,x^2+y^2+根号xy的最大值
答案是9/8

Answer 1

∵x>0y>0x+y=1
∴可设x=cos²a,y=sin²a,a∈(0,π/2)
∴u= x^2+y^2+√(xy)
=cos⁴a+sin⁴a+sinacosa
=(cos²a+sin²a)²-2sin²acos²a+sinacosa
=1-1/2(sin²2a)+1/2(sin2a)
设t=sin2a∈(0,1]
∴u=-1/2*t²+1/2*t+1
=-1/2(t²-t+1/4)+9/8
=-1/2(t-1/2)²+9/8
∴当 t=1/2 时,umax=9/8
即 x^2+y^2+√(xy)的最大值9/8