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.
Just one remark :
We have with the OP's constraints :
$$\left(\sum\limits_{j=1}^n \lambda_j x_j \right)\left(\sum\limits_{j=1}^n \lambda_j x_j^{-1} \right)\leq \left(\sum\limits_{j=1}^n \frac{1}{n} x_j \right)\left(\sum\limits_{j=1}^n \frac{1}{n} x_j^{-1} \right)\leq \dfrac{(x_1+x_n)^2}{4x_1x_n}.$$
With $\lambda_n\geq\cdots\geq\lambda_1$
Can you end now ?
A sketch for the LHS :
We start by prove the case $n=2$
The case $n=2$ have the form :
$$(\lambda_1a+(1-\lambda_1)b)(\lambda_2a+(1-\lambda_2)b)$$
Where $a+b=1$ .Remains to take the logarithm and use Jensen's inequality .
Now the case $n=3$
We have with $x_1\geq x_2\geq x_3>0$ and $\lambda_1\geq\lambda_2\geq\lambda_3>0$ and $\lambda_1+\lambda_2+\lambda_3=1$
$$\frac{4}{9}\frac{(\lambda_1x_1+(\lambda_2+\lambda_3)(x_2+x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2+\lambda_3}{x_2+x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})}\geq \frac{(\lambda_1x_1+\lambda_2x_2+\lambda_3x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2}{x_2}+\frac{\lambda_3}{x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})} $$
And :
$$\frac{(\lambda_1x_1+\frac{(1-\lambda_1)}{2}(x_2+x_3))(\frac{\lambda_1}{x_1}+\frac{(1-\lambda_1)}{2}\left(\frac{1}{x_2}+\frac{1}{x_3}\right))}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})}\geq \frac{(\lambda_1x_1+\lambda_2x_2+\lambda_3x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2}{x_2}+\frac{\lambda_3}{x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})} $$
This method works in the general case .
Best Answer
EDIT (2019-06-19): Here is a complete result.
Recall from the classical proof of the Cauchy-Schwarz-inequality, that there exists the equality $$ \sum_{i=1}^n \lambda_ix_i^2 - \frac{(\sum_{i=1}^n\lambda_ix_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} = \sum_{i=1}^n \lambda_i \left[ x_i - a y_i \right]^2 $$ with $$ a = \frac{\sum_{i=1}^n\lambda_ix_iy_i}{\sum_{i=1}^n\lambda_iy_i^2} $$
Hence the claim can be rewritten (introducing $F(x_i,y_i,\lambda_i)$) as: $$ \lambda_1 \le F(x_i,y_i,\lambda_i) =\sum_{i=1}^n \lambda_i \left[ x_i - a y_i \right]^2 + \left(\sum_{i=1}^n x_iy_i\right)^2 \le \lambda_n. $$
For the left inequality, we note that $0 \le \lambda_1 \le \lambda_i$ and $0 \le \lambda_1 \le 1$, hence it suffices to show: \begin{align} \lambda_1& \le \lambda_1 \sum_{i=1}^n \left[ x_i - a y_i \right]^2 + \lambda_1 \left(\sum_{i=1}^n x_iy_i\right)^2 \\ \Longleftrightarrow 1 &\le \sum_{i=1}^n \left[ x_i - a y_i \right]^2 + \left(\sum_{i=1}^n x_iy_i\right)^2 \\ &= \sum_{i=1}^n \left[ x_i^2 - 2 a x_iy_i + a^2y_i^2\right] + \left(\sum_{i=1}^n x_iy_i\right)^2 \\ &=1 + a^2 -2a \sum_{i=1}^n x_iy_i +\left(\sum_{i=1}^n x_iy_i\right)^2 \\ &=1 + (a - \sum_{i=1}^n x_iy_i )^2 \end{align} and this establishes the left inequality.
For the right inequality we do the following. Let $\sum_{i=1}^n x_iy_i = q$. Replace $x_i$ with $x_i = q y_i + n_i$. The reason to call the new variable $n_i$ is that $\sum_{i=1}^n y_i n_i = \sum_{i=1}^n y_i (x_i - q y_i) = q - q \sum_{i=1}^n y_i^2 = 0$, so the $(n_i)$ can be understood as the vector component of the vector $x$ which is normal (hence the n) to the $y$-vector. We have $\sum_{i=1}^n n_i^2 = \sum_{i=1}^n (x_i - q y_i)^2 = 1 - 2q^2 +q^2 = 1 - q^2$, which will be used below.
With this replacement, the expression in question becomes \begin{align} F(x_i,y_i,\lambda_i) &= \sum_{i=1}^n \lambda_i(q y_i + n_i)^2 + q^2 - \frac{(\sum_{i=1}^n\lambda_i(q y_i + n_i)y_i)^2}{\sum_{i=1}^n\lambda_iy_i^2}\\ &= q^2\sum_{i=1}^n \lambda_i y_i ^2 + 2q \sum_{i=1}^n \lambda_i y_i n_i + \sum_{i=1}^n \lambda_i n_i^2 + \\ &\qquad + q^2- \frac{(q \sum_{i=1}^n\lambda_i y_i^2 + \sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2}\\ &= q^2\sum_{i=1}^n \lambda_i y_i ^2 + 2q \sum_{i=1}^n \lambda_i y_i n_i + \sum_{i=1}^n \lambda_i n_i^2 + \\ &\qquad + q^2- q^2 \sum_{i=1}^n\lambda_i y_i^2 - 2q \sum_{i=1}^n\lambda_in_iy_i -\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2}\\ &= \sum_{i=1}^n \lambda_i n_i^2 + q^2 -\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \end{align} Structurally, this looks strikingly similar to the original formulation. The difference (which we will exploit) is that the vector $(n_i)$ has a relation to the $q$, which was not present before.
Replacing $q^2 = 1 - \sum_{i=1}^n n_i^2 $ (we had computed that already above) and bounding $\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \ge 0$ gives \begin{align} F(x_i,y_i,\lambda_i) &\le \sum_{i=1}^n \lambda_i n_i^2 + 1 - \sum_{i=1}^n n_i^2 \\ &\le \lambda_n \sum_{i=1}^n n_i^2 + 1 - \sum_{i=1}^n n_i^2 \\ &= 1 + (\lambda_n - 1)\sum_{i=1}^n n_i^2 \end{align} Further, we have that $\lambda_n - 1 \ge 0 $ and $\sum_{i=1}^n n_i^2 = 1 -q^2 \le 1$, so we can conclude $$ F(x_i,y_i,\lambda_i) \le 1 + (\lambda_n - 1) = \lambda_n $$ which is the desired result for the right inequality. This completes the proof. $\qquad \square$
Some interpretation: The bounding $\frac{(\sum_{i=1}^n\lambda_in_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \ge 0$ gives rise to the conclusion that this term is "not important" so it could be bounded away. This is the case, since without the $\lambda_i$, this term would become zero, as $\sum_{i=1}^n n_iy_i = 0$. Indeed, few computer simulations show that, for various choices of the $\lambda_i$, the maximum of the expression in question will be obtained when the vector $x$ is chosen almost perfectly perpendicular to the vector $y$, hence $x \simeq n$, which means that the discussed bounding is "save" as the term becomes small and thus doesn't produce much difference to the true result.