Proving the Kantorovich inequality by another inequality

Question

I wan to use the following inequality

Let $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$ be real numbers. Let $\frac{b_1}{a_1} = \max \{\frac{b_i}{a_i}, i=1,2, \ldots, n \}$ and $\frac{b_n}{a_n} = \min \{\frac{b_i}{a_i}, i=1,2, \ldots, n \}$. Then, this inequality holds $$\frac 1 2 \left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 \left(\sum_1^n a_i^2\right) ^2 \ge \left(\sum_1^n a_i^2\right) \left(\sum_1^n b_i^2\right)-\left(\sum_1^n a_i b_i\right)^2$$

to prove the Kantorovich inequality?. The Kantorovich inequality reads as

Let $\lambda_i>0$, $\sum_1^n \lambda_i =1$, $x_1=\max\{x_i, i=1, \cdots n\}$, and $x_n=\min\{x_i, i=1, \cdots n\}$. Then, the following holds $$\left(\sum_1^n \lambda_i x_i \right) \left(\sum_1^n \frac{\lambda_i}{x_i} \right) \le \left(\frac{x_1+x_n}{2\sqrt{x_1x_n}}\right)^2$$

Answer 1

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*}