Question

已知x>=0,y>=0,求证0.5(x+y)^2+0.25(x+y)>=x√y+y√x

Answer 1

(x+y)^2/2+(x+y)/4>=2*((x+y)^3/8)^(1/2)

(x+y)^2/2+(x+y)/4>=(x+y)*((x+y)/2)^(1/2)

(x+y)^2/2+(x+y)/4>=2(xy)^(1/2) *((x+y)/2)^(1/2)

(x+y)^2/2+(x+y)/4>=(2yx^2+2xy^2)^(1/2)

(x+y)^2/2+(x+y)/4>=(yx^2+xy^2+yx^2+xy^2)^(1/2)

(x+y)^2/2+(x+y)/4>=(yx^2+xy^2+2xy)^(1/2)

(x+y)^2/2+(x+y)/4>=((xy^(1/2)+yx^(1/2))^2)^(1/2)

(x+y)^2/2+(x+y)/4>=xy^(1/2)+yx^(1/2)