Prove that inequality Hardy inequality 
Suppose $n$ is a positive integer, $2n$ reals $x_i, y_i (1\le i \le n) $ satisfy 
  $$\sum_{i=1}^n x_i^2 = \sum_{i=1}^n y_i^2 = 1.$$
  positive reals $0 < \lambda_1 \le \lambda_2 \le ... \le \lambda_n, \ 1 \in [\lambda_1, \lambda_n].$ Prove that
  $$\lambda_1 \le \sum_{i=1}^n \lambda_ix_i^2 + \left(\sum_{i=1}^n x_iy_i\right)^2 - \frac{(\sum_{i=1}^n\lambda_ix_iy_i)^2}{\sum_{i=1}^n\lambda_iy_i^2} \le \lambda_n.$$

It seems stronger than Cauchy-Schwarz inequality.
$$\sum_{i=1}^{n}\lambda_{i}x^2_{i}\sum_{i=1}^{n}\lambda_{i}y^2_{i}\ge (\sum_{i=1}^{n}\lambda_{i}x_{i}y_{i})^2$$
Idear 2: I  try to use 
From Pólya-Szegö’s inequality, we have for $0 < m_1 \leqslant u_k \leqslant M_1$ and $0 < m_2 \leqslant v_k \leqslant M_2$, 
$$\left(\sum u_k^2 \right) \left( \sum v_k^2 \right) \leqslant \frac14 \left( \sqrt{\frac{M_1 M_2}{m_1m_2}} + \sqrt{\frac{m_1 m_2}{M_1 M_2}} \right)^2 \left( \sum u_k v_k\right)^2$$
But I can't it.Thanks
 A: 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.
