We say that matrices are orthogonal (in Frobenius inner product sense) if $\langle A, B \rangle_F = \text{tr}(A^HB) = 0$. Let $A, B$ be traceless matrices in $\mathbb{C}^{4 \times 4}$ that are orthogonal to each other. From this we have
$$
\lVert A + B \rVert_F^{2} = \text{tr}((A+B)^H (A+B)) = tr(A^H A) + tr(A^H B) + tr(B^H A) + tr(B^H B) = \lVert A \rVert^2_F + \lVert B \rVert^2_F
$$
Which is basically a Pythagorean theorem (or Pythagorean orthogonality).
Let $\lVert X \rVert_{2}$ mean spectral or matrix 2-norm which is equal to biggest singular value of $X$.
If traceless matrices in $\mathbb{C}^{4 \times 4}$ $A, B$ are orthogonal (in Frobenius inner product sense) can we say that $\lVert A + B \rVert_2^2 \leq \lVert A \rVert_2^2 + \lVert B \rVert_2^2$?
It is easy to get to
$$
\lVert A + B \rVert_2^2 \leq \lVert A \rVert_2^2 + \lVert B \rVert_2^2 + \max_{\lVert x \rVert_2 = 1}\lVert (A^H B + B^H A)x \rVert_2
$$
But I have hard time going further from here to prove or disprove this statement.
If the statement above is obviously false, then is following true:
$$
\lVert A + B \rVert_2^2 \leq \lVert A \rVert_F^2 + \lVert B \rVert_F^2
$$
Best Answer
The first inequality fails even for diagonal matrices. Indeed, let $$A=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix},\quad B= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$ Then $\langle A,B\rangle_F =0,$ $\|A\|_2^2=\|B\|_2^2=1$ and $\|A+B\|_2^2=4.$
Concerning the second statement $\|C\|_2\le \|C\|_F$ for any matrix $C.$ Hence $$\|A+B\|_2^2\le \|A+B\|_F^2=\|A\|_F^2+\|B\|_F^2$$ provided that $\langle A, B\rangle_F =0.$