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)的最小值是

Question

急用，谢谢

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

Answer 2

【注：因y²-z²=(y²-x²)+(x²-z²),故（y²-z²)/(z+x)=(y²-x²)/(z+x)+(x²-z²)/(z+x)】证明：易知，该式为对称式。不妨设z≥x≥y＞0.w=(z²-x²)/(x+y)+(x²-y²)/(y+z)+(y²-z²)/(z+x)=（z²-x²)/(x+y)+(x²-y²)/(y+z)-(x²-y²)/(z+x)-(z²-x²)/(z+x)=(z²-x²)[1/(x+y)-1/(z+x)]+(x²-y²)[1/(y+z)-1/(z+x)]=(z²-x²)(z-x)/[(x+y)(z+x)]+(x²-y²)(x-y)/[(y+z)(z+x)].由z≥x≥y＞0易知，w=(z²-x²)(z-x)/[(x+y)(z+x)]+(x²-y²)(x-y)/[(y+z)(z+x)]≥0,等号仅当x=y=z＞0时取得。