This is slightly too long to be comment, though this is not an answer:
Edit: I am providing a few revisions, but I still have not found a proof.
My previous comment was not quite correct. I "rearranged" the inequality by raising both sides of the inequality to certain powers, and I did not take into account that we do not know a priori whether any of the terms are larger or smaller than $1$. Depending on the size of each term (in particular, whether it is smaller or larger than $1$), exponentiating both sides could potentially reverse the inequality. I have returned the question to its original form as suggested by the OP, and I have added one more observation.
Let me first say that I have been working on this question for a bit, and though I have not yet resolved it, I have been having fun trying!
Now, to emphasize the dependence on $n$, let's set
$$
\alpha_n = \sum_{i=1}^n a_i \qquad \beta_n = \sum_{i=1}^n b_i, \qquad \sigma_n = \sum_{i=1}^n c_i,
$$
where $c_i = \sqrt{a_i b_i}$. Further, let's put
$$
A_n = \prod_{i=1}^n (a_i)^{a_i}, \qquad B_n = \prod_{i=1}^n (b_i)^{b_i}, \qquad S_n = \prod_{i=1}^n (c_i)^{c_i}.
$$
Our goal is to show:
\begin{equation}
(1) \hspace{1in} \left(\frac{A_n}{(\alpha_n)^{\alpha_n}}\right)^{\frac{1}{\alpha_n}} \cdot \left(\frac{B_n}{(\beta_n)^{\beta_n}} \right)^{\frac{1}{\beta_n}} \leq \left(\frac{S_n}{(\sigma_n)^{\sigma_n}}\right)^{\frac{2\sigma_n}{\alpha_n \beta_n}}
\end{equation}
A few pedestrian observations:
If $a_i = b_i$ for $i = 1, \dots , n$ (which forces $c_i = a_i = b_i$), then $A_n = B_n = S_n$, we also have $\alpha_n = \beta_n = \sigma_n$, and (1) holds in this case.
Note that $2c_i \leq a_i + b_i$ as $2xy \leq x^2 + y^2$ for all real numbers $x, y$. Hence, $2\sigma_n \leq \alpha_n + \beta_n$. Furthermore, Cauchy-Schwarz gives $\sigma_n^2 \leq \alpha_n \beta_n$. Both of these observations imply that $(\sigma_n + 1)^2 \leq (\alpha_n + 1)(\beta_n + 1)$.
I would imagine that with enough creativity, one may find a proof of the inequality involving convexity or a simple application of the AM-GM inequality (which I suppose is much the same thing!).
I have been unable to prove the inequality even in the case $n = 2$, when I assume $\alpha_n = \beta_n = 1$. I am not hopeful for a proof of the general case.
It may not be easy to directly use the inequality you mentioned at the beginning to prove the Kantorovich inequality. Here is the proof based on this paper, as I commented above.
We assume that $x_n>0$. Note that
\begin{align*}
\sum_{i=1}^n\lambda_i x_i\sum_{i=1}^n\lambda_i\frac{1}{x_i} &= -\sum_{i=1}^n\lambda_i\bigg(x_i -\sum_{i=1}^n\lambda_i x_i \bigg)\bigg(\frac{1}{x_i} -\sum_{i=1}^n\lambda_i \frac{1}{x_i} \bigg)+1\\
&=-\sum_{i=1}^n\sqrt{\lambda_i}\bigg(x_i -\sum_{i=1}^n\lambda_i x_i \bigg)\sqrt{\lambda_i}\bigg(\frac{1}{x_i} -\sum_{i=1}^n\lambda_i \frac{1}{x_i} \bigg)+1\\
&\le \sqrt{\sum_{i=1}^n \lambda_i\bigg(x_i -\sum_{i=1}^n\lambda_i x_i \bigg)^2 }\sqrt{\sum_{i=1}^n \lambda_i\bigg(\frac{1}{x_i} -\sum_{i=1}^n\lambda_i \frac{1}{x_i} \bigg)^2 }+1.
\end{align*}
Moreover (see also this question),
\begin{align*}
\sum_{i=1}^n \lambda_i\bigg(x_i -\sum_{i=1}^n\lambda_i x_i \bigg)^2 &= \min_{a\in \mathbb{R}}\sum_{i=1}^n \lambda_i(x_i-a)^2\\
&\le \sum_{i=1}^n \lambda_i\bigg(x_i -\frac{x_1+x_n}{2} \bigg)^2\\
&\le \frac{(x_1-x_n)^2}{4}.
\end{align*}
Similarly,
\begin{align*}
\sum_{i=1}^n \lambda_i\bigg(\frac{1}{x_i} -\sum_{i=1}^n\lambda_i \frac{1}{x_i} \bigg)^2 &\le \frac{\big(\frac{1}{x_1}-\frac{1}{x_n}\big)^2}{4}\\
&=\frac{(x_1-x_n)^2}{4x_1^2 x_n^2}.
\end{align*}
Therefore,
\begin{align*}
\sum_{i=1}^n\lambda_i x_i\sum_{i=1}^n\lambda_i\frac{1}{x_i} & \le \frac{(x_1-x_n)^2}{4x_1x_n}+1\\
&=\frac{(x_1+x_n)^2}{4x_1x_n}.
\end{align*}
Best Answer
The assertion is false. A counterexample with $n=2$ is $$ a_1=a_2=x_1=x_2=1,\quad b_1=\tfrac12,\, y_1=4,\, b_2=2,\, y_2=\tfrac14. $$ Suppose we make the change of variables $w_i = a_i^2x_i$ and $z_i = b_i^2y_i$; then the assertion becomes $$ \frac{\sum a_i^{-1}w_i}{\sum b_i^{-1}z_i} \le \frac{\sum w_i}{\sum z_i} = 1 \implies \frac{\sum a_iw_i}{\sum b_iz_i} \ge 1. $$ In this formulation, it's natural to consider the case $w_i=z_i=1$; it's easy now to see that the assertion is likely false if we take the set of $a_i$ to be closed under taking reciprocals and the set of $b_i$ to be closed under taking reciprocals.