I recently saw an inequality for real numbers known as Kantorovich inequality for real numbers. This says:

Suppose $x_1< x_2< \dots < x_n$ are positive real numbers and let let $\lambda_1, \lambda_2, \dots \lambda_n \geq 0$ with $\sum\limits_{j=1}^n \lambda_i=1$. Then

$$\left(\sum\limits_{j=1}^n \lambda_j x_j \right)\left(\sum\limits_{j=1}^n \lambda_j x_j^{-1} \right)\leq \dfrac{(x_1+x_n)^2}{4x_1x_n}.$$

My question is can we replace the strict inequality for $x_i$ by $``\leq"$? It seems so but I cannot prove it. Can anyone help me on this? By proving or by supporting documents?

Thanks in advance.

  • 2
    $\begingroup$ Follow the proof. I believe that equality holds when half of the terms are $ = x_1$ and the other half of the terms are $ = x_n$, meaning that with your strict condition, we do have $ < $ for $ n > 2$. $\endgroup$
    – Calvin Lin
    Jul 3, 2021 at 13:53
  • $\begingroup$ Thankyou@CalvinLin. Would you please elaborate? I feel you said, with the strict $<$ condition, the resultant inequality would be strict. $\endgroup$
    – pmun
    Jul 3, 2021 at 13:59
  • 1
    $\begingroup$ @pmun: I posted a solution following a probabilistic approach to your problem discussing equality issues as well here I hope you found it useful. This included your particular form of the Kantorovich Inequality. There are other interesting extensions that can be deduced also by probabilistic methods. $\endgroup$
    – Mittens
    Jul 5, 2021 at 3:11
  • $\begingroup$ @OliverDiaz Thank you for your kind response. I have seen your reference. $\endgroup$
    – pmun
    Jul 5, 2021 at 9:39

1 Answer 1


Just one remark :

We have with the OP's constraints :

$$\left(\sum\limits_{j=1}^n \lambda_j x_j \right)\left(\sum\limits_{j=1}^n \lambda_j x_j^{-1} \right)\leq \left(\sum\limits_{j=1}^n \frac{1}{n} x_j \right)\left(\sum\limits_{j=1}^n \frac{1}{n} x_j^{-1} \right)\leq \dfrac{(x_1+x_n)^2}{4x_1x_n}.$$

With $\lambda_n\geq\cdots\geq\lambda_1$ Can you end now ?

A sketch for the LHS :

We start by prove the case $n=2$

The case $n=2$ have the form :


Where $a+b=1$ .Remains to take the logarithm and use Jensen's inequality .

Now the case $n=3$

We have with $x_1\geq x_2\geq x_3>0$ and $\lambda_1\geq\lambda_2\geq\lambda_3>0$ and $\lambda_1+\lambda_2+\lambda_3=1$

$$\frac{4}{9}\frac{(\lambda_1x_1+(\lambda_2+\lambda_3)(x_2+x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2+\lambda_3}{x_2+x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})}\geq \frac{(\lambda_1x_1+\lambda_2x_2+\lambda_3x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2}{x_2}+\frac{\lambda_3}{x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})} $$

And :

$$\frac{(\lambda_1x_1+\frac{(1-\lambda_1)}{2}(x_2+x_3))(\frac{\lambda_1}{x_1}+\frac{(1-\lambda_1)}{2}\left(\frac{1}{x_2}+\frac{1}{x_3}\right))}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})}\geq \frac{(\lambda_1x_1+\lambda_2x_2+\lambda_3x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2}{x_2}+\frac{\lambda_3}{x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})} $$

This method works in the general case .

  • $\begingroup$ Yes@ErikSatie, I have proved it, although differently. But your hint will make things easier for me. $\endgroup$
    – pmun
    Jul 3, 2021 at 19:43

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