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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.

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    $\begingroup$ 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
    Commented 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
    Commented Jul 3, 2021 at 13:59
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    $\begingroup$ @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
    Commented Jul 5, 2021 at 3:11
  • $\begingroup$ @OliverDiaz Thank you for your kind response. I have seen your reference. $\endgroup$
    – pmun
    Commented Jul 5, 2021 at 9:39
  • $\begingroup$ Does this answer your question? Understanding Kantorovich's inequality $\endgroup$
    – Mittens
    Commented Nov 15, 2023 at 20:05

2 Answers 2

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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 :

$$(\lambda_1a+(1-\lambda_1)b)(\lambda_2a+(1-\lambda_2)b)$$

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 .

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  • $\begingroup$ Yes@ErikSatie, I have proved it, although differently. But your hint will make things easier for me. $\endgroup$
    – pmun
    Commented Jul 3, 2021 at 19:43
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Fixing the values of the $ \lambda $s, write $ f ( x _ 1 , \ldots , x _ n ) $ for the left-hand side and $ g ( x _ 1 , \ldots , x _ n ) $ for the right-hand side (although this value of $ g $ happens to depend only on $ x _ 1 $ and $ x _ n $). Because both expressions may be written using only (finitely many) additions, multiplications, and divisions (by nonzero values), these are both continuous functions (on $ ( 0 , \infty ) ^ n $ rather than on $ \mathbb R ^ n $ to avoid division by zero). Now, $ g - f \geq 0 $ on the open set $ U = \{ x _ 1 , \ldots , x _ n \; | \; x _ 1 < \cdots < x _ n \} $, and the closure of this set (in $ ( 0 , \infty ) ^ n $) is $ \overline U = \{ x _ 1 , \ldots , x _ n \; | \; x _ 1 \leq \cdots \leq x _ n \} $. In other words, $ ( g - f ) [ U ] \subseteq [ 0 , \infty ) $; so $ ( g - f ) [ \overline U ] \subseteq \operatorname { Cl } [ 0 , \infty ) = [ 0 , \infty ) $. That is, $ g - f \geq 0 $ on $ \overline U $, which is what you want.

Short version: If a weak inequality between continuous functions holds assuming some conditions given by strict (or weak) inequalities (or equations), then it will continue to hold if all of the inequalities in the hypothesis are weakened (as long as everything is still continuous there). (Also, if a strict inequality holds under such conditions, then the weakening of the conclusion will hold if all of the inequalities in the hypothesis are weakened, in the same way.)

Note that we can't replace $ 0 < x _ 1 $ by $ 0 \leq x _ 1 $, since then the expressions could be undefined. (However you could interpret both sides as $ \infty $ when $ x _ 1 = 0 $ and then the inequality would continue to hold.)

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