How to proof that $H^{1/2}XH^{1/2}=\left(\frac{2\lambda_i^{1/2}\lambda_j^{1/2}}{\lambda_i +\lambda_j} \right)\circ\left( \frac{1}{2}(HX+XH) \right)$ 
I've got this from Hiai and Kosaki paper, could someone explain to me how to derive $H^{1/2}XH^{1/2}=\left(\frac{2\lambda_i^{1/2}\lambda_j^{1/2}}{\lambda_i +\lambda_j}  \right)\circ\left( \frac{1}{2}(HX+XH) \right)$ (red-boxed)
Also, what the meaning of "thanks to the standard $2\times 2$ matrix trick, one can reduce a proof to the case $H=K.$" What trick is this?
Any help would be appreciated.
Thanks so much
 A: The trick is probably
$$
\begin{bmatrix} H&0\\0&K\end{bmatrix}^{1/2}
\begin{bmatrix} 0&X\\0&0\end{bmatrix}
\begin{bmatrix} H&0\\0&K\end{bmatrix}^{1/2},
$$
since its norm is $\|H^{1/2}XK^{1/2}\|$ and
$$
\begin{bmatrix} H&0\\0&K\end{bmatrix}
\begin{bmatrix} 0&X\\0&0\end{bmatrix}+
\begin{bmatrix} 0&X\\0&0\end{bmatrix}\begin{bmatrix} H&0\\0&K\end{bmatrix}
=\begin{bmatrix}0&HX+XK\\0&0\end{bmatrix}.
$$
As for the fomula, doing the $2\times 2$ case to avoid dots,
\begin{align}
H^{1/2}XH^{1/2}
&=\begin{bmatrix} \lambda_1^{1/2}&0\\0&\lambda_2^{1/2}\end{bmatrix}
\begin{bmatrix} x&y\\ z&w\end{bmatrix}
\begin{bmatrix} \lambda_1^{1/2}&0\\0&\lambda_2^{1/2}\end{bmatrix}
=\begin{bmatrix} \lambda_1 x&\lambda_1^{1/2}\lambda_2^{1/2}y\\\lambda_1^{1/2}\lambda_2^{1/2} z&\lambda_2 w\end{bmatrix}\\[0.3cm]
&=\begin{bmatrix} \lambda_1& \lambda_1^{1/2}\lambda_2^{1/2}\\ \lambda_1^{1/2}\lambda_2^{1/2}&\lambda_2\end{bmatrix} \circ X,
\end{align}
and
\begin{align}
XH+HX&=\begin{bmatrix} \lambda_1 x&\lambda_1 y\\\lambda_2 z&\lambda_2w\end{bmatrix}
+\begin{bmatrix} \lambda_1 x&\lambda_2 y\\\lambda_1 z&\lambda_2w\end{bmatrix}
=\begin{bmatrix} 2\lambda_1 x&(\lambda_1+\lambda_2) y\\(\lambda_1+\lambda_2 )z&2\lambda_2w\end{bmatrix}\\[0.3cm]
&=\begin{bmatrix} 2\lambda_1&\lambda_1+\lambda_2\\\lambda_1+\lambda_2&2\lambda_2\end{bmatrix}\circ X.
\end{align}
So, denoting the Hadamard inverse of $A$ by $A^\#$,
\begin{align}
H^{1/2}XH^{1/2}
&=\begin{bmatrix} \lambda_1& \lambda_1^{1/2}\lambda_2^{1/2}\\ \lambda_1^{1/2}\lambda_2^{1/2}&\lambda_2\end{bmatrix}\circ \begin{bmatrix}2\lambda_1&\lambda_1+\lambda_2\\\lambda_1+\lambda_2&2\lambda_2\end{bmatrix}^\#\circ (HX+XH)\\[0.3cm]
&=\begin{bmatrix} \lambda_1^{1/2}\lambda_1^{1/2}& \lambda_1^{1/2}\lambda_2^{1/2}\\ \lambda_1^{1/2}\lambda_2^{1/2}&\lambda_2^{1/2}\lambda_2^{1/2}\end{bmatrix}\circ \begin{bmatrix}\lambda_1+\lambda_1&\lambda_1+\lambda_2\\\lambda_1+\lambda_2&\lambda_2+\lambda_2\end{bmatrix}^\#\circ (HX+XH)\\[0.3cm]
&=\begin{bmatrix} \frac{\lambda_1^{1/2}\lambda_1^{1/2}}{\lambda_1+\lambda_1}& \frac{\lambda_1^{1/2}\lambda_2^{1/2}}{\lambda_1+\lambda_2}\\ \frac{\lambda_1^{1/2}\lambda_2^{1/2}}{\lambda_1+\lambda_2}&\frac{\lambda_2^{1/2}\lambda_2^{1/2}}{\lambda_2+\lambda_2}\end{bmatrix}\circ (HX+XH)
\end{align}
