matrices - Frobenius Norm, Triangle inequality, and complex conjugates

I found a thread that solved the problem I need to turn in (or confirmed that I had done it correctly) but doesn't really resolve some confusion I have regarding norms and inner products.

I need to show that the Frobenius norm obeys the general definition of a matrix norm, and only the triangle inequality is giving me any trouble, but that's been worked to death : Frobenius Norm Triangle Inequality

But I went about it somewhat differently and it's highlighted a few concepts I'm shaky on. Here is my approach:

Starting from the defintion
$$
||A||_F = \left( \sum^m_{i=1} \sum^n_{j=1} |A_{ij}|^2 \right)^{1/2}.
$$

Now consider

$$
||A+B||_F = \left( \sum^m_{i=1} \sum^n_{j=1} |A_{ij}+B_{ij}|^2 \right)^{1/2}.
$$

Noting that each element $A_{ij}, B_{ij}$ can be thought of as vectors in $\Re^2$, I can apply the good old fashioned triangle inequality to the square root of each summand

such that $|A_{ij}+B_{ij}|\leq |A_{ij}|+|B_{ij}|$

Squaring both sides gives me

$|A_{ij}+B_{ij}|^2\leq |A_{ij}|^2+|B_{ij}|^2 +2|A_{ij}||B_{ij}|$

I see that I'm on the right track, but I'm afraid I'm a bit stuck here.

If I sum over all elements, I get back

$||A+B||_F^2 \leq ||A||_F^2 + ||B||_F^2 + 2\sum^m_{i=1} \sum^n_{j=1}|A_{ij}||B_{ij}|$

Is my approach hopelessly flawed, or is there some way I can salvage this?