An upper bound of product of two inner products

The question is,

Let $$A \in M_{n \times n}(\Bbb C)$$ be a self-adjoint matrix. Arrange the eigenvalues of $$A$$ as $$0 < \lambda_1 \le \lambda_2 \le \cdots \le \lambda_n.$$

Show that $$\langle Ax, x \rangle \langle A^{-1}x,x \rangle \le \frac{(\lambda_1+\lambda_n)^2}{4 \lambda_1 \lambda_n}$$ for all $$||x||=1, x\in V$$

My approach:

Since $$A$$ is self-adjoint, $$V$$ has an orthonormal basis consisting of eigenvectors of $$A.$$ Let $$\{v_1,v_2, \cdots, v_n\}$$ be an orthonormal basis of $$V$$ where $$Av_i=\lambda_iv_i, \forall i \in \Bbb N.$$

Any arbitrary $$x \in V$$ with $$||x||=1$$ can be written as $$x=\sum_{i=1}^n c_iv_i$$ such that $$\sum_{i=1}^n|c_i|^2=1$$ where $$c_i \in \Bbb C, \forall i \in \Bbb N.$$ Therefore, we can write,

$$\langle Ax, x \rangle \langle A^{-1}x, x \rangle=\langle A(\sum_{i=1}^n{c_iv_i}),\sum_{i=1}^n{c_iv_i}\rangle \langle A^{-1}(\sum_{i=1}^n{c_iv_i}),\sum_{i=1}^n{c_iv_i}\rangle$$ $$=\left(\sum_{i=1}^n|c_i|^2\lambda_i\right)\left(\sum_{i=1}^n\frac{1}{\lambda_i}|c_i|^2 \right)$$

And I could not figure out how to proceed after this. I managed to show the inequality holds when $$A$$ is a $$2\times 2$$ matrix. Also clearly if $$c_1=c_n= \sqrt{\frac{1}{2}}, c_2=\cdots=c_{n-1}=0,$$ then the equality holds. But apart from these, I could not do much. I tried using Lagrange multiplier but haven't had much luck with that either. So any help would be really appreciated.

• This is sometimes called the Kantorovich inequality. Depending on the technical tools you have at your disposal, you may be able to find a proof that uses those tools via a web search. It's possible to give fairly elementary proofs. (Not trying to be unhelpful here; I just don't have time this afternoon to come up with or track down a proof and translate it into an answer.) Commented Jan 29, 2021 at 23:41
• For a algebraic proof see: math.stackexchange.com/a/4874794/468313 Commented Mar 4 at 5:52

Remark: As pointed out by @leslie townes, this is the so-called the Kantorovich inequality. Years ago, I came up with a proof myself (However, perhaps it is not new).

Problem: Let $$a_k \ge 0$$, $$k = 1, 2, \cdots, n$$ with $$\sum_{k=1}^n a_k = 1$$. Let $$0 < \lambda_1 \le \lambda_2 \le \cdots \le \lambda_n$$. Prove that $$\Big(\sum_{k=1}^n \lambda_k a_k\Big) \Big(\sum_{k=1}^n \frac{a_k}{\lambda_k}\Big)\le \frac{(\lambda_1+\lambda_n)^2}{4\lambda_1\lambda_n}.$$

Proof:

If $$\lambda_1 = \lambda_n$$, the inequality is clearly true. In the following, assume $$\lambda_1 < \lambda_n$$.

Consider the optimization problem $$\max_{a_k\ge 0, \forall k; \ \sum_{k=1}^n a_k = 1} \Big(\sum_{k=1}^n \lambda_k a_k\Big) \Big(\sum_{k=1}^n \frac{a_k}{\lambda_k}\Big).$$ Let $$(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$$ be a global maximizer.

We claim that if $$\lambda_1 < \lambda_k < \lambda_n$$, then $$a_k^\ast = 0$$. Indeed, if $$a_k^\ast > 0$$, let $$a_1' = a_1^\ast + (1 - y) a_k^\ast, \quad a_k' = 0, \quad a_n' = a_n^\ast + y a_k^\ast$$ where $$\frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1} < y < \frac{\lambda_n}{\lambda_k}\cdot \frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1}$$. We have $$a_1' + a_k' + a_n' = a_1^\ast + a_k^\ast + a_n^\ast,$$ and \begin{align} \lambda_1 a_1' + \lambda_k a_k' + \lambda_n a_n' &= \lambda_1 a_1^\ast + \lambda_n a_n^\ast + [\lambda_1 + (\lambda_n - \lambda_1) y]a_k^\ast \\ &> \lambda_1 a_1^\ast + \lambda_n a_n^\ast + \left(\lambda_1 + (\lambda_n - \lambda_1)\cdot \frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1}\right)a_k^\ast\\ &= \lambda_1 a_1^\ast + \lambda_n a_n^\ast + \lambda_k a_k^\ast, \end{align} and \begin{align} \frac{a_1'}{\lambda_1} + \frac{a_k'}{\lambda_k} + \frac{a_n'}{\lambda_n} &= \frac{a_1^\ast}{\lambda_1} + \frac{a_n^\ast}{\lambda_n} + \left(\frac{1}{\lambda_1} - \frac{\lambda_n - \lambda_1}{\lambda_1 \lambda_n}y\right)a_k^\ast \\ &> \frac{a_1^\ast}{\lambda_1} + \frac{a_n^\ast}{\lambda_n} + \left(\frac{1}{\lambda_1} - \frac{\lambda_n - \lambda_1}{\lambda_1 \lambda_n} \cdot \frac{\lambda_n}{\lambda_k}\cdot \frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1}\right)a_k^\ast \\ &= \frac{a_1^\ast}{\lambda_1} + \frac{a_n^\ast}{\lambda_n} + \frac{a_k^\ast}{\lambda_k}. \end{align} However, this contradicts the optimality of $$(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$$.

Now, assume that, among $$\lambda_1, \lambda_2, \cdots, \lambda_n$$, there are $$p$$ elements equal to $$\lambda_1$$, and there are $$q$$ elements equal to $$\lambda_n$$, where $$p + q \le n$$. Denote $$A = a_1^\ast + a_2^\ast + \cdots + a_p^\ast$$ and $$B = a_{n-q+1}^\ast + a_{n-q+2}^\ast + \cdots + a_n^\ast$$. We have $$A + B \le 1$$. We have \begin{align} &\Big(\sum_{k=1}^n \lambda_k a_k^\ast\Big) \Big(\sum_{k=1}^n \frac{a_k^\ast}{\lambda_k}\Big)\\ =\ & \Big(\lambda_1 A + \lambda_n B\Big)\left(\frac{A}{\lambda_1} + \frac{B}{\lambda_n}\right)\\ =\ & (A + B)^2 + \left(\frac{\lambda_1}{\lambda_n} + \frac{\lambda_n}{\lambda_1} - 2\right)AB\\ \le\ & 1 + \left(\frac{\lambda_1}{\lambda_n} + \frac{\lambda_n}{\lambda_1} - 2\right)\cdot \frac{1}{4}\\ =\ & \frac{(\lambda_1+\lambda_n)^2}{4\lambda_1\lambda_n}. \end{align} We are done.

• It is also a special case of the “discrete Pólya-Szegő inequality” which is proven in this answer: math.stackexchange.com/a/880365/42969. Commented Sep 29, 2022 at 12:52
• @MartinR Thank you! Commented Sep 29, 2022 at 13:46
• Simple and Clever. What was the motivation for discovering $a^*_k = 0$? Commented Nov 18, 2023 at 8:13
• @HoseinRahnama Thanks. You may investigate the case $n=3$ (e.g. by Lagrange Multiplier, or numerical example). Commented Nov 18, 2023 at 9:43