(1)证明:由已知,4Sn=
a2n+2an,且an>0.
当n=1时,4a1=
a21+2a1,解得a1=2.
当n≥2时,有4Sn−1=
a2n−1+2an−1.
于是4Sn−4Sn−1=
a2n−
a2n−1+2an−2an−1,即4an=
a2n−
a2n−1+2an−2an−1.
于是
a2n−
a2n−1=2an+2an−1,即(an+an-1)(an-an-1)=2(an+an-1).
∵an+an-1>0,
∴an-an-1=2(n≥2).
故数列{an}是首项为2,公差为2的等差数列,且an=2n;
(2)证明:∵an=2n,则[1
Sn=
1
n(n+1)=
1/n−
1
n+1],
∴[1
S1+
1
S2+…+
1
Sn=(1−
1/2)+(
1
2−
1
3)+…+(
1
n−
1
n+1)=1−
1
n+1<1.
∵1−
1
n+1]随着n的增大而增大,
∴当n=1时取最小值[1/2].
故原不等式成立;
(3)由