Question

已知实数m,n满足m≥0,n≥0且m+n=1,则m²/(m+2)+n²/(n+2)的最小值为

Answer 1

把n=1-m代入，原式=m²/(m+2)+(1-m)²/(3-m)=m²/(m+2)-(m-1)²/(m-3)
=[m²(m-3)-(m+2)(m-1)²]/[(m+2)(m-3)]
化简=-2/[(m+2)(m-3)],因m≥0,n≥0且m+n=1得0≤m≤1
得(m+2)(m-3)<0,
得(m+2)(m-3)绝对值越大，化简式-2/[(m+2)(m-3)]越小
(m+2)(m-3)=m²-m-6=(m-1/2)²-6-1/4,得m=1/2时最小，
即原式最小值为0.32。

Answer 2

把n=1-m代入，原式=m²/(m+2)+(1-m)²/(3-m)=m²/(m+2)-(m-1)²/(m-3)
=[m²(m-3)-(m+2)(m-1)²]/[(m+2)(m-3)]
化简=-2/[(m+2)(m-3)],因m≥0,n≥0且m+n=1得0≤m≤1
得最小值为(m+2)(m-3)