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.
Best Answer
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.