Question

x,y,z为正实数,则( z^2-x^2)/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)的

Answer 1

设x≥y≥z

所以x^2≥y^2≥z^2≥0

1/(y+z)≥1/(x+z)≥1/(x+y)

所以x^2/(x+y)+y^2/(y+z)+z^2/(x+z)(乱序和)

≤x^2/(y+z)+y^2/(x+z)+z^2/(x+y)(顺序和)

左边的移到右边去

[x^2/(y+z)-y^2/(y+z)]+[y^2/(x+z)-z^2/(x+z)]+[z^2/(x+y)-x^2/(x+y)]≥0

中括号里的合并

所以(z^2-x^2)/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)≥0

所以最小值是0