Let $$ J = \begin{bmatrix} 0 & 1 & \dots & 1\\ 1 & 0 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 0 & \dots & 0 \end{bmatrix}, \ \ \ \ \ K= \begin{bmatrix} 0 & 0 & \dots & 0\\ 0 & k_{11} & \dots & k_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & k_{n1} & \dots & k_{nn} \end{bmatrix} $$ two $N \times N$ matrices and F = J+K.
Doing some numerical experiments I noticed that the following relation holds: $$ \frac{\lVert F \rVert_F}{\lVert F \rVert} \leq \frac{\lVert K \rVert_F}{\lVert K \rVert} + \frac{\lVert J \rVert_F}{\lVert J \rVert} $$ where $\lVert . \rVert_F$ is the Frobenius norm and $\lVert . \rVert$ is the spectral norm.
Is it true for all K? How can I prove it?
Yes, it's true and the result is rather trivial. In general, suppose $$ J=\pmatrix{0&u^T\\ v&0},\,K=\pmatrix{0&0\\ 0&B}\text{ and }F=J+K=\pmatrix{0&u^T\\ v&B}. $$ Then $$ \|J\|_2=\max\{\|u\|_2,\|v\|_2\}=\max\{\|Fe_1\|_2,\|e_1^TF\|_2\}\le\|F\|_2 $$ and $\|K\|_2=\|B\|_2\le\|F\|_2$ because $B$ is a submatrix of $F$. Therefore $$ \frac{\|K\|_F}{\|K\|_2} + \frac{\|J\|_F}{\|J\|_2} \ge\frac{\|K\|_F+\|J\|_F}{\|F\|_2} \ge\frac{\sqrt{\|K\|_F^2+\|J\|_F^2}}{\|F\|_2} =\frac{\|F\|_F}{\|F\|_2}. $$