Decomposition of a $2 \times 2$ matrix into the difference of 2 positive semi-definite matrices. For the matrix $$A = \begin{bmatrix}
    0      & 1 \\
    1    & 0 \\
\end{bmatrix}$$ with norm $||A|| = \sum_{i,j = 1,2} |a_{ij}|$
Show any decomposition of $A = C - B$ with $B, C$ being positive semi-definite matrices, satisfies
$||B|| + ||C|| \geq 4$.
I've started with setting the conditions $c_{11} = b_{11}, c_{22} = b_{22}, c_{ij} = 1 + b_{ij} $ off the diagonal.
I then worked through $x^T A x = x^T(C-B)x$, with each term $\geq 0$, and worked through that while applying the above conditions, however I just arrive at a trivial answer $2 x_1 x_2 = 2x_1 x_2$, so I'm unsure how to proceed with this problem.

We are given A=C-B with :
$$ A = \begin{bmatrix}
    0      & 1 \\
    1    & 0 \\
\end{bmatrix} $$
Let us take B & evaluate C :
$$ B = \begin{bmatrix}
    a      & b \\
    b    & c \\
\end{bmatrix} $$
$$ C = A+B = \begin{bmatrix}
    a      & b+1 \\
    b+1    & c \\
\end{bmatrix} $$
With $ B $ & $ C $ Positive Semi-Definite, we get (XXX) :
$ ac-b^2 \ge 0 $
$ ac-(b+1)^2 \ge 0 $
$ ac \ge b^2 $
$ ac \ge (b+1)^2 $
Addition gives $ 2ac \ge b^2 + (b+1)^2 $ &
$ 2ac \ge b^2 + (b+1)^2 $
Minimum value of RHS is at $ b=-1/2 $ , where $ b^2 + (b+1)^2 = 1/4 + 1/4 = 1/2 $ , which gives :
$ 2ac \ge 1/2 $
$ |2a| |2c| \ge 1 $
Standard InEquality gives :
$ |2a| + |2c| \ge $
$ |a| + |c| \ge 1 $ (XXX)
It is easy to see that (YYY) :
$ ||B|| = |a|+|c|+|b|+|b| $ &
$ ||C|| = |a|+|c|+|b+1|+|b+1| $
$ ||B||+||C|| = ( |a|+|c|+|b|+|b| ) + ( |a|+|c|+|b+1|+|b+1| ) $
$ ||B||+||C|| = 2( |a|+|c| ) + 2( |b|+|b+1| ) $ (YYY)
Use (XXX) in (YYY) to get the minimum value 4.
$ ||B||+||C|| \ge 2( 1 ) + 2( 1 ) $
$ ||B||+||C|| \ge 4 $
