Question

已知两个正项数列{An}和{Bn}满足an+bn=n-1求证根号An+1/2+根号bn+1/ 2<=根号2n

已知两个正项数列{An}和{Bn}满足an+bn=n-1求证根号An+1/2+根号bn+1/ 2<=根号2n

Answer 1

an+bn=n-1
(an+1/2)+(bn+1/2)=n
[√(an+1/2)]^2+[√(bn+1/2)]^2=n
[√(an+1/2)+√(bn+1/2)]^2-2√(an+1/2)√(bn+1/2)=n
[√(an+1/2)+√(bn+1/2)]^2=n+2√(an+1/2)√(bn+1/2)
≤n+[√(an+1/2)]^2+[√(bn+1/2)]^2
=n+(an+1/2)+(bn+1/2)
=n+an+bn+1
=2n
即
[√(an+1/2)+√(bn+1/2)]^2≤2n
所以
√(an+1/2)+√(bn+1/2)≤√(2n)。

Answer 2

用A[n]表示第n项
A[n]+B[n]=n-1
A[n]+1/2+B[n]+1/2=n①(考虑到A[n]和B[n]为正，用a^2+b^2≥2ab)
n=A[n]+1/2+B[n]+1/2≥2√(A[n]+1/2)√(B[n]+1/2)②
①+②:2n≥A[n]+1/2+B[n]+1/2+2√(A[n]+1/2)√(B[n]+1/2)
=(√(A[n]+1/2)+√(B[n]+1/2))^2
∴√(A[n]+1/2)+√(B[n]+1/2)≤√(2n).