李江云
幼苗
共回答了16个问题采纳率：87.5% 举报
n^2+(a-4)n+3+a≥0,n是正整数,
∴a(n+1)≥-(n^2-4n+3),
∴a≥-(n^2-4n+3)/(n+1),
设u=n+1,则u≥2,n=u-1,
n^2-4n+3=(u-1)^2-4(u-1)+3=u^2-6u+8,
(n^2-4n+3)/(n+1)=u+8/u-6≥4√2-6,但u≠2√2,
∴u=3时u+8/u-6取最小值17/3-6=-1/3,
∴-(n^2-4n+3)/(n+1)的最大值=1/3,
∴a≥1/3,为所求.
1年前
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