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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:

$$\left(\sum_1^n \lambda_i x_i \right) \left(\sum_1^n \frac{\lambda_i}{a_i} \right) \le \left(\frac{x_1+x_n}{2\sqrt{x_1x_n}}\right)^2$$

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    $\begingroup$ Please include all relevant information to the problem (e.g. hypotheses and uncommon definitions) in the actual post. Do not simply post links to them if you can help it. I.e. please take the time to write what the Kantorovich inequality is here in this post. Also take the time to display the inequality you wish to take as hypothesis here instead of linking to another question for it. Noone wants to flip back and forth between three or four pages just to read a question. $\endgroup$ – JMoravitz Aug 26 '16 at 18:26
  • $\begingroup$ See this paper for a proof. $\endgroup$ – Gordon Aug 26 '16 at 20:41
  • $\begingroup$ @JMoravitz thank to You, I edited my question $\endgroup$ – Oai Thanh Đào Aug 27 '16 at 0:52
  • $\begingroup$ @Gordon thank to dear, I edited my question $\endgroup$ – Oai Thanh Đào Aug 27 '16 at 0:53
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    $\begingroup$ Changed the right side $\frac{x_1+x_n}{2\sqrt{x_1x_n}}$ of the last inequality to $\left(\frac{x_1+x_n}{2\sqrt{x_1x_n}}\right)^2$. $\endgroup$ – Gordon Aug 28 '16 at 17:51
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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*}

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