Can proof the Kantorovich inequality by a inequality Could you let me know can using the inequality as following:
Let $a_1, a_2,\ldots,a_n, b_1, b_2,\ldots,b_n$, let $\frac{b_1}{a_1} = \max \{\frac{b_i}{a_i}, i=1,2, \ldots, n \}$,
$\frac{b_n}{a_n} = \min \{\frac{b_i}{a_i}, i=1,2, \ldots, n \}$  show that:
$$\frac 1 2 \left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 \left(\sum_1^n a_i^2\right) ^2 \ge \left(\sum_1^n a_i^2\right) \left(\sum_1^n b_i^2\right)-\left(\sum_1^n a_i b_i\right)^2$$
to proof the Kantorovich inequality?. The Kantorovich inequality as following:
Let $\lambda_i>0$, and $\sum_1^n \lambda_i =1$, $x_1=\max\{x_i, i=1, \cdots n\}$, $x_n=\min\{x_i, i=1, \cdots n\}$ then:
$$\left(\sum_1^n \lambda_i x_i \right) \left(\sum_1^n \frac{\lambda_i}{a_i}  \right) \le \left(\frac{x_1+x_n}{2\sqrt{x_1x_n}}\right)^2$$
 A: 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*}
