Question

求证:a^2+b^2+2>=2a+2b; 已知xy>0,求xy+1/xy+y/x+x/y>=4

Answer 1

求证:a^2+b^2+2>=2a+2b;

a^2+b^2+2-(2a+2b)

=(a^2-2a+1)+(b^2-2b+1)

=(a-1)^2+(b-1)^2≥0

所以a^2+b^2+2≥2a+2b

已知xy>0,求xy+1/xy+y/x+x/y>=4

xy>0,

1/xy>0,

x/y>0,

y/x>0

xy+1/xy>=2根号[xy*1/xy]=2

x/y+y/x>=2根号[x/y*y/x]=2

所以xy+1/xy+y/x+x/y>=4