Prove that if $A\ge B$ then $\left[ {\begin{array}{*{20}{c}} A & B \\ B & A \\ \end{array}} \right]\ge0$. Let $A$ and $B$ be $n \times n$ matrices, i.e., $A, B \in  M_n$. Also, $A \ge 0$, $B \ge 0$, and $A-B \ge 0$ which mean all these matrices are semi-positive-definite.
Why does 
$\left[ {\begin{array}{*{20}{c}}
   A & B  \\
   B & A  \\
\end{array}} \right] \ge 0$?

First, write
$$
\begin{pmatrix}
x' & y'
\end{pmatrix}
\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
=x'Ax+y'Bx+x'By+y'Ay.
$$
Next, consider the RHS above
\begin{align*}
\text{RHS}&=x'(A-B)x+x'Bx+y'Bx+x'By+y'By+y'(A-B)y\\
&=x'(A-B)x+(x+y)'Bx+(x+y)'By+y'(A-B)y\\
&=x'(A-B)x+(x+y)'B(x+y)+y'(A-B)y
\end{align*}
which is nonnegative due to the positive semidefiniteness of $A-B$ and $B$. The claim follows.
