# Kantorovich inequality and Cauchy-Schwarz inequality

On the wikipedia site for the Kantorovich inequality, it is claimed

... the Kantorovich inequality is a particular case of the Cauchy–Schwarz inequality...

Here, "Kantorovich inequality" refers to $$(x^\top A \, x) \, (x^\top A^{-1} \, x) \le \frac{(m+M)^2}{4 \, m \, M} \, \|x\|^4$$ for a symmetric, positive definite matrix $$A$$ and $$m$$, $$M$$ denote the smallest and largest eigenvalue of $$A$$.

I was wondering what is meant by the above claim of wikipedia. Is it really true that the Kantorovich inequality is a particular case of CSI (in the sense of: "can be easily derived from")?

The closest assertion I was able to find is from this paper: If we plug in $$x = \sqrt{A} \, y$$, we arrive at $$\| A \, y\|^2 \, \|y\|^2 \le \frac{(m+M)^2}{4 \, m \, M} \, (y^\top A \, y)^2$$ which is a reverse of the special case $$y^\top A \, y \le \|A \, y\| \, \|y\|$$ of CSI.

• Side remark: in Pták's proof (published in American Mathematical Monthly), Kantrovich's inequality is an easy consequence of the AM-GM inequality. Jul 3, 2021 at 19:56

A recent related posting ask for conditions on equality in the Kantorovich inequality. I think is more appropriate to have a solution in this (older) posting and address the question here and the issue of equality.

The approach here is a probabilistic generalization of the result of the OP.

Theorem: Suppose $$X$$ is a random variable on $$(\Omega,\mathscr{F},\mathbb{P})$$ taking values in an interval $$[a,b]$$ with $$0. Then, \begin{align} 1\leq \mathbb{E}[X]\mathbb{E}[X^{-1}]\leq \frac{(a+b)^2}{4ab} \tag{0}\label{kan}\end{align} Equality on the left-hand-side happens iff $$X$$ is constant $$\mathbb{P}$$-a.s; equality on the right-handside happens iff $$\mathbb{P}[X=a]=\mathbb{P}[X=b]=\frac12$$.

Proof: Since $$\phi(x)=\frac{1}{x}$$ is convex in $$(0,\infty)$$, the left-side inequality follows from Jensen's inequality. Equality iff either $$\phi(X)$$ is linear $$\mathbb{P}$$-a.s. or $$X$$ constant a.s.

The right-side inequality follows from the Cauchy-Schwartz inequality. First notice that for any part of square integrable random variables $$X,Y$$ $$\mathbb{E}\big[(X-\mathbb{E}(X))(Y-\mathbb{E}[Y])\big]\leq\big(E\big[(X-\mathbb{E}(X))^2\big]\big)^{1/2}\big(E\big[(Y-\mathbb{E}(Y))^2\big]\big)^{1/2}$$ Equality iff either one of the random variables is a constant $$\mathbb{P}$$-a.s., or if there is $$c>0$$ such that $$|X-\mathbb{E}[X]|=c|Y-\mathbb{E}[Y]|$$. This is a well known result in Probability: the covariance of two random variables is at most the product of the variances of the random variables.

Taking $$Y=X^{-1}$$, we get $$\mathbb{E}[X]\mathbb{E}[X^{-1}]\leq 1+\Big(E\big[\big(X-\mathbb{E}[X]\big)^2\big]\Big)^{1/2}\Big(E\big[\big(X^{-1}-\mathbb{E}[X^{-1}]\big)^2\big]\Big)^{1/2}$$

Since $$0, we also have that $$0<\frac1b\leq\frac{1}{X}\leq \frac1a$$. Recall that the mean $$E[Z]$$ of any square integrable random variable $$Z$$ satisfies $$E\big[(Z-\mathbb{E}[Z])^2\big]=\inf_{a\in\mathbb{R}}E\big[(Z-a)^2\big]$$. In particular, \begin{align} \mathbb{E}\big[\big(X-\mathbb{E}[X]\big)^2\big]&\leq \mathbb{E}\big[\big(X-\frac{a+b}{2}\big)^2\big]\leq\Big(\frac{b-a}{2}\Big)^2\\ E\big[\big(X^{-1}-\mathbb{E}[X^{-1}]\big)^2\big]&\leq \mathbb{E}\big[\big(X^{-1}-\frac{a^{-1}+b^{-1}}{2}\big)^2\big]\leq\Big(\frac{b-a}{2ab}\Big)^2 \end{align} Putting things together, we obtain $$\mathbb{E}[X]\mathbb{E}[X^{-1}] \leq 1+\frac{b-a}{2}\frac{b-a}{2ab}=\frac{(a+b)^2}{4ab}$$

The conditions for equality in the theorem follow by considering the cases where equality occur in Jensen's inequality and in Cauchy-Schwartz inequality. I leave that for anybody interested in verifying the conditions.

The inequality in the OP is a particular case of the Theorem outlined above. First notice that since $$A$$ is positive definite matrix, we may assume, without loss of generality, that $$A$$ is a diagonal matrix with positive entries $$m=\lambda_1\leq\ldots\leq \lambda_n=M$$. Furthermore, it is enough to consider $$x$$ with $$\|x\|^2_2=1=\sum^n_{k=1}|x_k|^2$$. Define $$\Omega=\{1,\ldots,k\}$$, $$\mathscr{F}$$ to collection of all subsets of $$\Omega$$, and $$\mathbb{P}[\{k\}]=|x_k|^2$$, and $$X(k)=\lambda_k$$.

A survey with other interesting extensions can be found in Bühler, W. Two proofs of the Kantorovich Inequality and generalizations, Revista Colombiana de Matemáticas, Vol. 21 (1987), pp.147-154.

• This is really a nice and beautiful proof! Are you aware of any proof based on direct optimization of $(x^\top A \, x) \, (x^\top A^{-1} \, x) / (x^{\top} x)^2$ Nov 17, 2023 at 11:57
• @HoseinRahnama: I am not aware of a proof that uses direct optimization techniques instead of using Cauchy-Shwartz inequality (in some form). Here a present a more general generalization of the Kantorovich inequality, still with a probabilistic flavor, and using simple Calculus optimization techniques. Nov 17, 2023 at 15:06
• What you commented is also fine, but I think the use of Cauchy-Schwartz in a way which I pointed out is easier to follow. :) Nov 17, 2023 at 16:00
• @HoseinRahnama: Either way, as long as one get the 1 plus the product of standard variations ($Y=X^{-1})$) the rest follows through. The other posting I linked in my comments has probably a more direct link to the business of optimization. Nov 17, 2023 at 16:04
• @HoseinRahnama: This may be of interest to you Raghavachari, M., A linear programing proof of Kantorovich's inequality. The American StatIstIcIan, 40 (1986) 136-137. Nov 17, 2023 at 18:15