类似地,若 an≤an+1a_n\le a_{n+1}an≤an+1,则 an+2≤an2+2an+1≤an+12+2an+1+1=an+1+1a_{n+2}\le\sqrt{a_n^2+2a_{n+1}}\le\sqrt{a_{n+1}^2+2a_{n+1}+1}=a_{n+1}+1an+2≤an2+2an+1≤an+12+2an+1+1=an+1+1;将第一条引理中 an,an+1a_n,a_{n+1}an,an+1 的角色互换做同样的计算,可直接得到 an+2≤an+1a_{n+2}\le a_{n+1}an+2≤an+1。