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

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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, 2016 at 18:26
  • $\begingroup$ See this paper for a proof. $\endgroup$
    – Gordon
    Aug 26, 2016 at 20:41
  • $\begingroup$ @JMoravitz thank to You, I edited my question $\endgroup$ Aug 27, 2016 at 0:52
  • $\begingroup$ @Gordon thank to dear, I edited my question $\endgroup$ Aug 27, 2016 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, 2016 at 17:51

1 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*}


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