How prove this $\frac{1}{b_{k+1}b_{k+2}}+\frac{1}{b_{k+2}b_{k+3}}+\cdots+\frac{1}{b_{2k}b_{2k+1}}>\frac{1}{12345},$ let sequence $\{b_{n}\}$,and $b_{n}>0$,let
$$S_{n}=b_{1}+b_{2}+\cdots+b_{n}\le n^{\frac{3}{2}},\forall n\ge 1$$
show that
$$\dfrac{1}{b_{k+1}b_{k+2}}+\dfrac{1}{b_{k+2}b_{k+3}}+\cdots+\dfrac{1}{b_{2k}b_{2k+1}}>\dfrac{1}{12345},k\ge 1$$
I think this problem maybe
$$b_{n}=S_{n}-S_{n-1}?$$
But this problem I can't it.Thank you 

Since the function $\frac{1}{x^2}$ is a convex function for $x > 0$, for any $k$ positive numbers $x_i, i = 1,\ldots,k$, we can apply Jensen's inequality to conclude
$$\frac{1}{k} \sum_{i=1}^k\frac{1}{x_i^2} 
\ge \frac{1}{\left(\frac{1}{k} \sum\limits_{i=1}^k x_i\right)^2}
\quad\implies\quad
\sum_{i=1}^k\frac{1}{x_i^2} \ge \frac{k^3}{\left(\sum\limits_{i=1}^k x_i\right)^2}
$$
Now take $\;\displaystyle x_i = \frac{b_{k+i}+b_{k+i+1}}{2}$ for $i = 1,\ldots, k$
and notice 
$$\sum_{i=1}^k x_i = \frac12\left(S_{2k+1}+S_{2k} - S_{k+1} - S_{k}\right) \le S_{2k+1} \le (2k+1)^{3/2}$$
We obtain
$$\sum_{i=1}^k\frac{1}{b_{k+i}b_{k+i+1}}
\ge \sum_{i=1}^k\frac{1}{x_i^2}
\ge \frac{k^3}{\left(\sum\limits_{i=1}^n x_i\right)^2} 
\ge \frac{k^3}{S_{2k+1}^2} 
\ge \left(\frac{k}{2k+1}\right)^3
\ge \frac{1}{27} > \frac{1}{12345}
$$
because $k \ge 1$.
