An inequality for positive definite matrix with trace 1 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 $f$.

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 $f(x)=\frac{(n - 1)^{2n}}{n^n}$.
