4 reworded the previous edit to make it clear which parts of question were new

The most appealing statement of the Bessis-Moussa-Villani conjecture is as follows:

Conjecture: For all Hermitian positive semidefinite $n\times n$ matrices $A$ and $B$, and all positive integer $m$, the polynomial function $$t \in \mathbb{R}\mapsto g(t) \equiv tr[(A + t B)^m] = \sum\limits_{ k=0}^m a_kt^k$$ has only nonnegative coefficients $a_k, k=1,\cdots,m$.

Most recent and past research concerns on the quantities $m$ and $n$. What about trying to prove the conjecture in the following way: first show that $a_m\ge0$, $a_0\ge0$, $a_{m-1}\ge0$, $a_1\ge0$, $a_{m-2}\ge0$, $a_2\ge0$. And then go on to show $a_3\ge 0$ (and so is $a_{m-3}\ge 0$)...

But it seems difficult to show $a_3\ge 0$. Can anyone share some idea on this particular coefficient?

UPDATED What is the largest term in $S_{2m,m}(AB)$ ? More precisely, consider the word in two positive definite letters $A^{k_1}BA^{k_2}B\cdots A^{k_m}B$ , where $(k1,k2,\cdots,k_m)$ is a pair of nonnegative integer solution of $k_1+k_2+\cdots+k_m=m$ . Is it true that $tr(A^{k_1}BA^{k_2}B\cdots A^{k_m}B)\le tr(A^mB^m)$ ?

3 update a problem.

The most appealing statement of the Bessis-Moussa-Villani conjecture is as follows:

Conjecture: For all Hermitian positive semidefinite $n\times n$ matrices $A$ and $B$, and all positive integer $m$, the polynomial function $$t \in \mathbb{R}\mapsto g(t) \equiv tr[(A + t B)^m] = \sum\limits_{ k=0}^m a_kt^k$$ has only nonnegative coefficients $a_k, k=1,\cdots,m$.

Most recent and past research concerns on the quantities $m$ and $n$.

What about trying to prove is the conjecture largest term in the following way: first show that $a_m\ge0$, $a_0\ge0$, $a_{m-1}\ge0$, $a_1\ge0$, $a_{m-2}\ge0$, S_{2m,m}(AB)$? More precisely, consider the word in two positive definite letters$a_2\ge0$. And then go on to show A^{k_1}BA^{k_2}B\cdots A^{k_m}B$ , where $a_3\ge 0$ (and so (k1,k2,\cdots,k_m)$is a pair of nonnegative integer solution of$a_{m-3}\ge 0$)..k_1+k_2+\cdots+k_m=m$ .

But Is it seems difficult to show true that $a_3\ge 0$. Can anyone share some idea on this particular coefficienttr(A^{k_1}BA^{k_2}B\cdots A^{k_m}B)\le tr(A^mB^m)$? 2 retag; fixed some of the English; tweaked title to be more descriptive # AnunmaturestrategyforWhat'sknownaboutthe3rdcoefficientinthe BMV conjecture? The most appealing statement of BMV the Bessis-Moussa-Villani conjecture is as follows: Conjecture: For all Hermitian positive semidefinite$n\times n$matrices$A$and$B$, and all positive integer$m$, the polynomial function$t $t \in \mathbb{R}\mapsto g(t) \equiv tr[(A + t B)^m] = \sum\limits_{ k=0}^m a_kt^k$ a_kt^k has only nonnegative coefficients $a_k, k=1,\cdots,m$.

Most recent and past research concerns on the quantities $m$ and $n$. What about trying to prove the conjecture this in the following way, (for any $m$ and $n$, it is obvious to see) : first show that $a_m\ge0$, $a_0\ge0$, $a_{m-1}\ge0$, $a_1\ge0$, $a_{m-2}\ge0$, $a_2\ge0$. And then go on to show $a_3\ge 0$ (and so is $a_{m-3}\ge 0$)...

But it seems difficult to show $a_3\ge 0$. Can anyone share some idea on this particular coefficient?

1