An inequality for positive definite matrix with trace 1

Question

Given a positive definite matrix $A \in \mathbb R^{n \times n}$. If $\operatorname{trace}(A) = 1$, for $n \geq 3$, prove that $$\text{det}(A) \leq \frac{n^n}{(n-1)^{2n}} \text{det}(I -A)^2$$ and the equation only holds when $A = \frac{1}{n}I$.

What I have tried: For any positive definite matrix $P \in \mathbb{R}^{n \times n}$, there exists an invertible matrix $\Omega$, s.t. \begin{align} &P = \Omega^{-1} \Sigma \Omega \\ &\Sigma = \text{diag}(x_1, \cdots, x_n) \\ &x_1, \cdots, x_n > 0 \end{align} We have $\text{det}(P) = \text{det}(\Sigma) =\prod_{i = 1}^n x_i$ and $\text{det}(I - P) = \text{det}(I - \Sigma) = \prod_{i = 1}^n (1 - x_i)$. Therefore, we only need to prove $\forall x_1, \cdots, x_n$ with $x_1, \cdots, x_n > 0$ and $\sum_{i=1}^n x_i=1$, the inequality holds: \begin{align} \prod_{i = 1}^n (1 - x_i)^2 \geq \frac{(n - 1)^{2n}}{n^n} \prod_{i = 1}^n x_i \end{align} Assume $f(x_1, \cdots, x_n) = \prod_{i = 1}^n (1 - x_i)^2 / \prod_{i = 1}^n x_i$ and we then find the minimum of $f$.

Answer 1

In order to investigate behavior of the function $f$, at the first we consider the following auxiliary problem.

Let $0<x,y,a$ and $x+y+a=1$. If $a$ is fixed then

$$h(x,y)=\frac{(1-x)^2(1-y)^2}{xy}=\frac{(a+xy)^2}{xy}=\frac{a^2}{xy}+2a+xy.$$

A function $\frac{a^2}{t}+t$ has a derivative $-\frac{a^2}{t^2}+1$, so it decreases when $0<t<a$ and increases when $a<t<1$. By AM-GM inequality, $xy\le \frac{(1-a)^2}4$ and the equality is attained iff $x=y$. In particular, if $\frac{(1-a)^2}4\le a$ then $h(x,y)$ attains its minimum at $(x,y)$ iff $x=y$. It is easy to check that $\frac{(1-a)^2}4\le a$ iff $$a\ge 3-2\sqrt{2}=0.1715\dots.$$

Consider the function $f(x_1,\dots, x_n)$. Fixing the smallest $x_l$ we restrict domain of $f$ to a compact set $C$ consisting of $x$ such that $\{x_j\ge x_l\}$ for each $j$ and $\sum x_j=1$. Since $f$ on $C$ is continuous, it attains its minimum at some point $x$. Let $x_i$ be the largest coordinate of $x$. Then $$x_j+x_k\le \frac 23(x_i+x_j+x_k)\le \frac 23<1-(3-2\sqrt{2})$$ for each remaining distinct $j$ and $k$, so the minimality of $f(x)$ on $C$ and the properties of the function $h$ imply that $x_j=x_k=t$ for each remaining $j$ and $k$. Then $x_i=1-(n-1)t$ and

$$f(x)=r(t)=\frac{((n-1)t)^2}{1-(n-1)t}\left(\frac{(1-t)^2}{t}\right)^{n-1}=\frac{(n-1)^2(1-t)^{2n-2}}{(1-(n-1)t)t^{n-3}}.$$

If $t$ tends to $0$ or to $\frac 1{n-1}$ then $r(t)$ tends to infinity. So $r(t)$ attains its minimum at a compact subset of an interval $(0, \frac 1{n-1})$ in some its point $s$. Then $r’(s)=0$. This easily implies that

$$\left((1-s)^{2n-2}\right)’ (1-(n-1)s)s^{n-3}=(1-s)^{2n-2}\left((1-(n-1)s)s^{n-3}\right)'$$

$$(2-2n)(1-(n-1)s)s^{n-3}=(1-s)\left((1-(n-1)s)s^{n-3}\right)'$$

The following cases are possible

1) $n=3$. Then $-4(1-2s)=(1-s)\left(1-2s\right)'$ and $s=\frac 13$.

2) $n>3$. Then

$$(2-2n)(1-(n-1)s)s^{n-3}=(1-s)\left((n-3)s^{n-4}-(n-1)(n-2)s^{n-3}\right)$$

$$(2-2n)(1-(n-1)s)s=(1-s)\left((n-3)-(n-1)(n-2)s\right)$$

$$s^2(n^2-n)+s(n^2-4n+1)-n+3=0$$

$$s=\frac 1n\mbox{ or }s=-\frac {n-3}{n-1}<0.$$

Thus anyway $s=\frac 1n $. Then all $x_j$ equal to $\frac 1n$ and $f(x)=\frac{(n - 1)^{2n}}{n^n}$.

Answer 2

Thanks so much for the help from @Alex Ravsky. I also write one proof based on his ideas.

When $n = 3$, since $x_1 + x_2 + x_3 = 1$, at least one of $x_1$, $x_2$, $x_3$ is larger than $\frac{1}{3}$. Without loss of generality, we assume $x_1 \geq \frac{1}{3}$. Then construct the function $h(x_2, x_3)$ for $x_2, x_3 > 0$ and $x_2 + x_3 = 1 - x_1$: \begin{align} h(x_2, x_3) = \frac{(1 - x_2)^2 (1 - x_3)^2}{x_2 x_3} = \frac{x_1^2}{x_2 x_3} + x_2 x_3 + 2 x_1 \end{align} Using the mean value inequality, we have $x_2 x_3 \leq \frac{(1 - x_1)^2}{4}$. Thus if $\frac{(1 - x_1)^2}{4} \leq x_1$, i.e. $x_1 \geq 3 - 2 \sqrt 2$, then we have \begin{align} h(x_2, x_3) \geq \frac{4 x_1^2}{(1 - x_1)^2} + \frac{(1 - x_1)^2}{4} + 2 x_1 \end{align} and the equality holds iff $x_2 = x_3 = \frac{1 - x_1}{2}$. Since $x_1 \geq \frac{1}{3} > 3 - 2 \sqrt 2$, the condition holds. Now consider the function $f(x_1, x_2, x_3)$. Then \begin{align} f(x_1, x_2, x_3) = \frac{(1 - x_1)^2}{x1} h(x_2, x_3) \geq 4 x_1 + \frac{(1 - x_1)^4}{4 x_1} + 2 (1 - x_1)^2 \end{align} and the equality holds iff $x_2 = x_3 = \frac{1 - x_1}{2}$ for every $x_1 \geq \frac{1}{3}$. Assume $p(x_1) = 4 x_1 + \frac{(1 - x_1)^4}{4 x_1} + 2 (1 - x_1)^2$, then for $x_1 \geq \frac{1}{3}$, we have \begin{align} p'(x_1) = \frac{(1 + x)^3 (-1 + 3 x)}{4 x^2} \geq 0, \end{align} thus $p(x_1)$ attains it minimum iff $x_1 = \frac{1}{3}$, i.e. \begin{align} f(x_1, x_2, x_3) \geq \left(\frac{(1 - \frac{1}{3})^2}{\frac{1}{3}}\right)^3 = \frac{(3 - 1)^{2 \cdot 3}}{3^3} \end{align} and the equality holds iff $x_1 = \frac{1}{3}$ and $x_2 = x_3 = \frac{1 - x_1}{2} = \frac{1}{3}$.

Then we consider the case for $n > 3$. Similarly, without loss of generality, we can assume $x_1 \geq \frac{1}{n}$. We still first fix $x_1$ and define \begin{align} h(x_2, \cdots, x_n) = \prod_{i=2}^n \frac{(1 - x_i)^2}{x_i}. \end{align} We then assume $x_n$ is the largest one in $x_2, \cdots, x_n$. For $\sum_{i \neq 1, n} x_i = 1 - x_1 - x_n$, we define the following functions ($1 < j, k < n$, $j \neq k$): \begin{align} g_{j, k}(x_j, x_k) = \frac{(1 - x_j)^2 (1 - x_k)^2}{x_j x_k} = \frac{a^2}{x_j x_k} + x_j x_k + 2 a, \end{align} where $a = \sum_{i \neq j, k} x_i$. Similarly, if $a \geq 3 - 2 \sqrt 2$, then $g_{j, k}$ attains its minimum iff $x_j = x_k$. If there exists $j$ and $k$ s.t. $a = \sum_{i \neq j, k} x_i < 3 - 2\sqrt 2$, i.e. $x_j + x_k > 2\sqrt 2 -2 > 0.82$. Then at least one of $x_j$ and $x_k$ is no less than $0.41$, e.g. $x_j$, then $x_n \geq x_j \geq 0.41$ and thus $a > x_n \geq 0.41$, which is a contradiction. Therefore all $\sum_{i \neq j, k} x_i \geq 3 - 2 \sqrt 2$. Now consider the function \begin{align} g(x_2, \cdots, x_{n-1}) = \prod_{i=2}^{n-1} \frac{(1 - x_i)^2}{x_i}. \end{align} We first only let $x_2$ and $x_3$ move and fix the others. To attain the minimum we only need to consider $g_{2, 3}(x_2, x_3)$ and it must be $x_2 = x_3 = t$ for any other fixed variables. Then let $x_4$ move and we need to consider $g_{3, 4}(t, x_4)$. To attain the minimum, we then have $x_3 = x_4 = t$. Repeatedly, we get $x_2 = \cdots = x_{n-1} = t$, thus $t = \frac{1 - x_1 - x_n}{n-2}$. Now consider $h(x_2, \cdots, x_n)$, we have \begin{align} h(x_2, \cdots, x_n) \geq \frac{(1 - x_n)^2}{x_n} \left( \frac{\left(1 - \frac{1 - x_1 - x_n}{n-2}\right)^2}{\frac{1 - x_1 - x_n}{n-2}} \right)^{n-2} \end{align} %Denote the right side as $p_{x_1}(x_n)$ and it can be proved that $p_{x_1}(x_n)$ attains the minimum when $x_n = \frac{1 - x_1}{n - 1}$ if $x_1 \leq \frac{1}{n}$ ($x_n \leq 1 - \frac{1}{n}$). Since $t = \frac{1 - x_1 - x_n}{n - 2}$, $x_n = 1 - x_1 - (n - 2) t$. Therefore we can consider the function \begin{align} r (t) = \frac{(x_1 +(n - 2) t)^2}{1 - x_1 - (n - 2) t} \left(\frac{(1 - t)^2}{t} \right)^{n-2} \end{align} Since $x_1 \geq \frac{1}{n}$, we have $x_n \leq 1 - \frac{1}{n}$. It can be proved that $r(t)$ attains the minimum at $t = \frac{1 - x_1}{n - 1}$. The proof is nothing than straightforward but complicated calculation thus we omit the proof. Therefore, $x_n = \frac{1 - x_1}{n - 1}$ and $x_2 = \cdots = x_{n-1} = \frac{1 - x_1}{n - 1}$. Eventually we have \begin{align} f(x_1, \cdots, x_n) \geq \frac{(1 - x_1)^2}{x_1} \left( \frac{\left(1 - \frac{1 - x_1}{n - 1}\right)^2}{\frac{1 - x_1}{n - 1}} \right)^{n-1} \end{align} Denote the right side as $q(x_1)$, then we have \begin{align} q'(x_1) = \frac{(n - 1) (1 - x_1)^2 \left(\frac{(n - 1) \left(1 - \frac{1 - x_1}{n - 1}\right)^2}{(1 - x_1)} \right)^n (n - x_1 - 2) (n x_1 - 1) }{(x_1^2 (n + x_1 - 2)^3} \end{align} Since $x_1 \geq \frac{1}{n}$, we have $1 - x_1 > 0$, $n + x_1 - 2 > 0$, $n - x_1 - 2 > 0$, $n x_1 - 1 \geq 0$. Thus $q'(x_1) \geq 0$ and $q(x_1)$ attains its minimum at $x_1 = \frac{1}{n}$, thus $x_2 = \cdots, x_n = \frac{1 - x_1}{n - 1}=\frac{1}{n}$. In other words, we prove that \begin{align} f(x_1, \cdots, x_n) \geq \left(\frac{(1 - \frac{1}{n})^2}{\frac{1}{n}}\right)^n = \frac{(n - 1)^{2 \cdot n}}{n^n} \end{align} and the equality holds iff $x_1 = \cdots x_n = \frac{1}{n}$.