若对于任意实数n属于正整数n^2+(a-4)n+3+a>=0,则实数a的取值范围
人气:360 ℃ 时间:2020-03-29 11:58:08
解答
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...
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