若x.y是正数,则（x+2y）^2+(y+2x)^2的最小值是?

楼主的意思是求(x+1/(2y))^2+(y+1/(2x))^2的最小值吧。(x+1/2y)^2+(y+1/2x)^2=x^2+x/y+1/(4y^2)+y^2+y/x+1/(4x^2)=[x^2+1/(4x^2)]+[y^2+1/(4y^2)]+(x/y+y/x)>=2sqrt(x^2 * 1/(4x^2))+2sqrt(y^2+1/(4y^2))+2sqrt(x/y * y/x)=1+1+2=4不等式取等当且仅当x^2=1/(4x^2),y^2=1/(4y^2),x/y=y/x,即x=y=sqrt(2)/2时故当x=y=sqrt(2)/2时（x+2y）^2+(y+2x)^2取最小值4