Prove $\sum\limits_{k=1}^n a_k\sum\limits_{k=1}^n \frac1{a_k}\le\left(n+\frac12\right)^2$ then $\max a_k \le 4\min a_k$ 
Prove that
if
  \begin{align}
&0<a_1,a_2,\dots,a_n \in \mathbb R,
&\left(\sum_{k=1}^n a_k\right)\left(\sum_{k=1}^n \frac1{a_k}\right)\le\left(n+\frac12\right)^2\\
\end{align}
then
  $$\max_k \space a_k \le 4\times\min_k\space a_k$$

What I've tried was
$$\left(\sum_{k=1}^n a_k\right)\left(\sum_{k=1}^n \frac1{a_k}\right)=n+\sum_{i\ne j}\frac{a_i}{a_j}=n+\sum_{i< j}\left(\frac{a_i}{a_j}+\frac{a_j}{a_i}\right)$$
which didn't help at all. Thanks.

Hint:
\begin{align*}
\min_j a_j &\leq a_k \quad \text{ for every } 1 \leq k \leq n\\
\sum_{k = 1}^n \min_j a_j &\leq \sum_{k = 1}^n a_k\\
n \, \min_j a_j &\leq \sum_{k = 1}^n a_k
\end{align*}
Similarly
$$
n \frac{1}{\max\limits_j a_j} \leq \sum_{k = 1}^n \frac{1}{a_k}
$$
Hence
$$
n^2 \frac{\min\limits_j a_j}{\max\limits_i a_i} \leq \left( \sum_{k = 1}^n a_k \right)\left( \sum_{k = 1}^n \frac{1}{a_k} \right) \leq \left( n + \frac{1}{2} \right)^2 \leq \ldots
$$
This should lead to something.
Also, obviously $\frac{\max\limits_i a_i}{\min\limits_j a_j} \geq 1$. (I like having different indeces $i, j$ but these are of course dummy variables)
