### Improvements of some Berezin radius inequalities

The Berezin transform $\widetilde{A}$ and the Berezin radius of an operator $A$ on the reproducing kernel Hilbert space over some set $Q$ with normalized reproducing kernel $k_{\eta}:=\dfrac{K_{\eta}}{\left\Vert K_{\eta}\right\Vert}$ are defined, respectively, by $\widetilde{A}(\eta)=\left\langle {A}k_{\eta},k_{\eta}\right\rangle$, $\eta\in Q$ and $\mathrm{ber} (A):=\sup_{\eta\in Q}\left\vert \widetilde{A}{(\eta)}\right\vert$. A simple comparison of these properties produces the inequalities
more » ... frac{1}{4}\left\Vert A^{\ast}A+AA^{\ast}\right\Vert \leq\mathrm{ber}^{2}\left( A\right) \leq\dfrac{1}{2}\left\Vert A^{\ast}A+AA^{\ast}\right\Vert $. In this research, we investigate other inequalities that are related to them. In particular, for$A\in\mathcal{L}\left( \mathcal{H}\left(Q\right) \right) $we prove that$\mathrm{ber}^{2}\left( A\right) \leq\dfrac{1}{2}\left\Vert A^{\ast}A+AA^{\ast}\right\Vert _{\mathrm{ber}}-\dfrac{1}{4}\inf_{\eta\in Q}\left(\left( \widetilde{\left\vert A\right\vert }\left( \eta\right)\right)-\left( \widetilde{\left\vert A^{\ast}\right\vert }\left( \eta\right)\right) \right) ^{2}.\$