$a_n=\dfrac 14n^2+n+\dfrac {7+(-1)^n}{8}$

因为数列 $\{a_n\}$ 为单调递增数列，$a_1=2>0$，所以 $a_n>0(n\in \mathbb N^*)$．
由题意得$$\begin{split}&2a_{2n}=a_{2n-1}+a_{2n+1},\\&a_{2n+1}^2=a_{2n}a_{2n+2},\end{split}$$于是$$2a_{2n}=\sqrt {a_{2n-2}a_{2n }}+\sqrt{a_{2n}a_{2n+2}},$$化简得$$2\sqrt{a_{2n}}=\sqrt {a_{2n-2} }+\sqrt{ a_{2n+2}},$$所以数列 $\{\sqrt {a_{2n}}\}$ 为等差数列．
又$$a_3=2a_2-a_1=6 , a_4=\dfrac {a_3^2}{a_2}=9,$$所以数列 $\{\sqrt {a_{2n}}\}$ 的首项为 $\sqrt{a_2}=2$，公差为 $d=\sqrt {a_4}-\sqrt{a_2}=1$，所以$$\sqrt{a_{2n}}=n+1,$$从而$$a_{2n}=(n+1)^2.$$结合$$ a_{2n-1}^2=a_{2n-2}a_{2n},$$可得$$a_{2n-1}=n(n+1).$$因此，当 $n$ 为偶数时$$a_n=\dfrac 14(n+2)^2,$$当 $n$ 为奇数时$$a_n=\dfrac {(n+1)(n+3)}{4}.$$所以数列 $\{a_n\}$ 的通项公式为\[\begin{split}a_n&=\dfrac 12[1+(-1)^{n+1}]\cdot \dfrac {(n+1)(n+3)}{4}+\dfrac 12[1+(-1)^{n }]\cdot \dfrac {(n+2)^2 }{4}\\&=\dfrac 14n^2+n+\dfrac {7+(-1)^n}{8}.\end{split}\]

略

因为\[\begin{split}a_n&= \dfrac 14n^2+n+\dfrac {7+(-1)^n}{8} \\ & \leqslant \dfrac 14n^2+n+1 \\&=\dfrac {(n+2)^2 }{4}\\&<\dfrac {1}{4}(n+2)(n+3),\end{split}\]所以$$\dfrac {1}{a_n}>\dfrac {4}{(n+2)(n+3)}=4\left(\dfrac 1{n+2}-\dfrac 1{n+3}\right),$$故\[\begin{split}S_n&=\dfrac {1}{a_1}+\dfrac {1}{a_2}+\dfrac {1}{a_3}+\cdots +\dfrac {1}{a_n}\\&>4\left[\left(\dfrac 13-\dfrac 14\right)+\left(\dfrac 14-\dfrac 15\right)+\cdots +\left(\dfrac 1{n+1}-\dfrac 1{n+2}\right)+\left(\dfrac 1{n+2}-\dfrac 1{n+3}\right) \right]\\&=4\left(\dfrac 13-\dfrac 1{n+3}\right)=\dfrac {4n}{3(n+3)}.\end{split}\]因此$$S_n>\dfrac {4n}{3(n+3)}, n\in \mathbb N^*.$$