  The Berezin symbol $\widetilde{A}$ of an operator $A$ acting on the
reproducing kernel Hilbert space ${\mathscr H}={\mathscr H(}\Omega)$ over some
(non-empty) set is defined by $\widetilde{A}(\lambda)=\langle
A\hat{k}_{\lambda},\hat{k}_{\lambda}\rangle\,\,\,(\lambda\in\Omega)$, where
$\hat{k}_{\lambda}=\frac{{k}_{\lambda}}{\|{k}_{\lambda}\|}$ is the normalized
reproducing kernel of ${\mathscr H}$. The Berezin number of operator $A$ is
defined by $\mathbf{ber}(A) = \underset{\lambda \in \Omega}{\sup}
\big|\tilde{A}(\lambda)\big|=\underset{\lambda \in \Omega }{\sup} \big|\langle
A\hat{k}_{\lambda}, \hat{k}_{\lambda}\rangle\big|$. Moreover
$\mathbf{ber}(A)\leqslant w(A)$ (numerical radius). In this paper, we present
some Berezin number inequalities. Among other inequalities, it is shown that if
  $\mathbf{T}=\left[\begin{array}{cc}
  A&B
  C&D
  \end{array}\right]\in {\mathbb B}({\mathscr H(\Omega_1)}\oplus{\mathscr
H(\Omega_2)})$, then
  \begin{align*}
  \mathbf{ber}(\mathbf{T}) \leqslant\frac{1}{2}\left( \mathbf{ber}(A)+
\mathbf{ber}(D)\right)+\frac{1}{2}\sqrt{\left( \mathbf{ber}(A)-
\mathbf{ber}(D)\right)^2+(\|B\|+\|C\|)^2}.
  \end{align*}
