an2+
=an2+Sn2 n2
[na1+1 n2
n(n-1)d]21 2
=an2+[a1+
(n-1)d]2,1 2
令
(n-1)d=t,1 2
an2+
=(a1+2t)2+(a1+t)2Sn2 n2
=2a12+6ta1+5t2
=5(t-
)2+2a12-3a1
5
,9a12
5
当t=
时,取到最小值3a1
5
即
(n-1)d=1 2
,即n=3a1
5
+1,6a1
5d
∵不等式an2+
≥ma12对任意等差数列{an}及任意正整数n都成立,Sn2 n2
∴m≤
.1 5
∴实数m的最大值为
.1 5
故选:D.