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.
