Under the given condition, for all $n\in \mathbb{N}$, stronger inequality holds: $\left(\sum_{i=1}^{n} k_ia_i \right)\left(\sum_{i=1}^{n} \frac{k_i}{a_i} \right) \le \frac{1}{4}\left(a+\frac{1}{a}\right)^2$ Proof: From the condition, we have: $(a-a_i)\left(a-\frac{1}{a_i} \right)\ge 0 \Leftrightarrow a_i+\frac{1}{a_i}\le a+\frac{1}{a} $ Now, from am-gm, it follows: $\left(\sum_{i=1}^{n} k_ia_i \right)\left(\sum_{i=1}^{n} \frac{k_i}{a_i} \right) \le \frac{1}{4}\left[ \sum_{i=1}^{n} k_i\left(a_i+\frac{1}{a_i}\right) \right]^2 \le \frac{1}{4}\left[ \sum_{i=1}^{n} k_i\left(a+\frac{1}{a}\right) \right]^2 = \frac{1}{4}\left(a+\frac{1}{a}\right)^2$

Another way to get $ a_i+\frac{1}{a_i}\le a+\frac{1}{a} $ is to notice we have $a\leq a_i\leq 1$ and that $f(x) = x+\frac {1} {x}$ is decreasing on $(0,1]$. Also, for good measure, we have by Cauchy-Schwarz the reverse inequality $\left(\sum_{i=1}^{n} k_ia_i \right)\left(\sum_{i=1}^{n} \frac{k_i}{a_i} \right) \geq \left (\sum_{i=1}^{n} k_i\right )^2 = 1$.