Interesting series

By the Cauchy-Schwarz Inequality,

$\begin{aligned} b_n &= \sum_{i = 1}^n a_n \sum_{i = 1}^n \dfrac{1}{a_n} \\ &= \sum_{i = 1}^n \sqrt{a_n}^2 \sum_{i = 1}^n \left ( \dfrac{1}{\sqrt{a_n}} \right)^2 \\ &\geq \left ( \sum_{i = 1}^n \sqrt{a_n} \dfrac{1}{\sqrt{a_n}} \right )^2 \\ &= n^2. \end{aligned}$

Thus, $\left \lfloor \sqrt{b_n} \right \rfloor \geq \left \lfloor \sqrt{n^2} \right \rfloor = n.$ From this, we can see that $\left \lfloor \sqrt{b_n} \right \rfloor$ and $\left \lfloor \sqrt{b_{n + 1}} \right \rfloor$ differ by at least one, so we conclude that yes , all the $\left \lfloor \sqrt{b_k} \right \rfloor$ are different.

It is enough to prove that for each $k$ $\sqrt{b_n}+1\leq\sqrt{b_{n+1}}$

Let $d=a_1+a_2+\dots+a_n, c=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dots+\dfrac{1}{a_n}$ and $x=a_{n+1}$ .

Now we have to prove: $\sqrt{(d+x)\left(c+\dfrac{1}{x}\right)}\geq\sqrt{cd}+1$ From that $\dfrac{d}{x}+cx\geq2\sqrt{cd}$ which is true because the relationship between arithmetic mean and geometric mean.