real analysis - A System Involving the Trigamma Function

Given $K \geq 2$ real numbers $a_1, \dots, a_K$, with $a_k > 0$ for $k=1,\dots,K$, consider the system of equations

\begin{equation}
(a_k - x_k) \psi^{(1)}(x_k) = \psi^{(1)} \left(\sum_{k=1}^{K} x_k \right) \sum_{k=1}^{K} (a_k - x_k), \quad (k=1,\dots,K),
\end{equation}

in the real variables $x_1,\dots,x_K$, with $x_k > 0$ for $k=1,\dots,K$,
where $\psi^{(1)}(x)$ is the trigamma function.

Is is true that the only solution of this system is $x_k=a_k$ for all $k=1,\dots,K$?

Any help is welcome.

Answer

The answer is simpler than I thought. Actually, it only uses the fact the two following properties of the trigamma function that immediately follow from its series representation:

(i) $\psi^{(1)}(x) > 0$ for any $x > 0$;

(ii) the map $(0,\infty) \ni x \mapsto \psi^{(1)}(x)$ is strictly decreasing.

Now, assume that $x_k \neq a_k$ for some $k$. Since $(x_1,\dots,x_k)$ is a solution of the system, we see that all the quantities $a_k - x_k$, for $k=1,\dots,K$, have the same sign. Then we would have, by setting $w_k= |a_k - x_k|$, and $W= \sum_{k=1}^{K} w_k$:

\begin{equation}
\psi^{(1)} \left(\sum_{k=1}^{K} x_k \right) = \frac{1}{W} \sum_{k=1}^{K} w_k \psi^{(1)}(x_k) > \min \{ \psi^{(1)}(x_k): k=1,\dots, K \} > \psi^{(1)} \left(\sum_{k=1}^{K} x_k \right),
\end{equation}

a contradiction.

QED