I am struggling for a while in proving the positive-semidefiniteness
( in the usual sense, i.e., $A$ is Hermitian and $x^*Ax\geq 0\,$ for all
$\,x \in \mathbb{C}^2\,\big)$
of the following matrix
$$A :=\,BC^\top + CB^\top+BB^\top,$$
under the hypotheses that both $B$ and $C$ have real entries and are positive-semidefinite in $\mathbb R^2 $: in the sense that
$$x^\top B x\geq 0\quad\forall\, x\in\mathbb{R}^2$$
and likewise for $C$. Furthermore, $B,C$ are not necessarily symmetric.
Numerical tests suggest that this holds true whenever the corresponding non-diagonal elements of $B$ and $C$ share the same sign, that is, if we set $B=\begin{bmatrix} b_1 & b_2\\b_3 & b_4\end{bmatrix}$ and $C=\begin{bmatrix} c_1 & c_2\\c_3 & c_4\end{bmatrix}$ then $sign(b_2)=sign(c_2)$ and $sign(b_3)=sign(c_3)$.
My most promising attempt has been effectively computing matrix $A$ and checking the principal minors. We have that
$$A=\begin{bmatrix}
2(l_1u_1+l_2u_2)+l_1^2+l_2^2 & l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4\\
l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4 & 2(l_3u_3+l_4u_4)+l_3^2+l_4^2
\end{bmatrix},$$
and the non-negativity of the diagonal elements of $A$ comes from the shared sign of the off-diagonal elements of $B$ and $C$. It remains to ensure that $\det(A)\geq 0$. My bet is on the following computations:
\begin{equation*}
\begin{aligned}
\det(A)&=4(b_1c_1+b_2c_2(b_3c_3+b_4c_4)+2b_1^2(b_3c_3+b_4c_4)+2b_2^2(b_3c_3+b_4c_4)+2b_3^2(b_1c_1+b_2c_2)+2b_4^2(b_1c_1+b_2c_2)\\
&\quad +b_1^2b_3^2+b_1^2b_4^2+b_2^2b_3^2+b_2^2b_4^2-(b_1c_3+b_3c_1+b_4c_2+b_2c_4+b_1b_3+b_2b_4)^2\\
&=4(b_1c_1+b_2c_2)(b_3c_3+b_4c_4)+2b_1^2({b_3c_3}+b_4c_4)+2b_2^2(b_3c_3+{b_4c_4})+2b_3^2({b_1c_1}+b_2c_2)+2b_4^2(b_1c_1+{b_2c_2})\\
&\quad +{b_1^2b_3^2}+b_1^2b_4^2+b_2^2b_3^2+{b_2^2b_4^2}-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-{b_1^2b_3^2}-{b_2^2b_4^2}\\
&\quad-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-{2b_1^2c_3b_3}-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4-{2b_3^2b_1c_1}-2b_3c_1b_2b_4\\
&\quad-2b_4c_2b_2c_4-2b_4c_2b_1b_3-{2b_4^2c_2b_2}-2b_2c_4b_1b_3-{2b_2^2c_4b_2}-2b_1b_3b_2b_4\\
&=4b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+4b_2c_2b_4c_4+2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1+b_1^2b_4^2+b_2^2b_3^2\\
&\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad -2b_3c_1b_2b_4-2b_4c_2b_2c_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3-2b_1b_3b_2b_4\\
&=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4+\underline{2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1}+\underline{(b_1b_4-b_2b_3)^2}\\
&\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad \underline{-2b_1c_3b_2b_4-2b_3c_1b_2b_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3}\\
&=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2\\
&\quad -2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad +(b_1b_4-b_2b_3)(b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3),
\end{aligned}
\end{equation*}
where I have underlined the terms involved in the final equality in order to facilitate the reading.
And now I am stuck. I don't know if there is any obscure equality to gather the remaining terms, and can not even ensure that
$$b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3\geq 0,$$
although intuition tells me it should be true.
If you have any ideas for the follow up from here, I would be happy to hear them.
Best Answer
In case $B$ and $C$ are positive semidefinite in mathematicians' definition, they are symmetric and $A=X^2-Y^2$, where $X:=B+C\succeq Y:=C\succeq0$. However, it is known that $X\succeq Y\succeq0$ in general does not imply that $X^2\succeq Y^2$ (see here for instance). Therefore $A$ is not always positive semidefinite.