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

题目

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

答案

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...