Fisher's Inequality states that if $v\ge k$, then $b\ge v$. In this case $k=4$.
I am still pretty new to designs, and so don't understand things fully yet. There is a formula for $b$ as follows:
$$b = \frac{\lambda v(v-1)}{k(k-1)}$$
Using this, I reduced the problem to showing that $\lambda(v-1)\ge 12$
I am not sure where to go with this now.
Fisher's inequality works under condition $v>k$. So I will assume that also $v>k$ in your case. Since one of the basic arguments leads to equality $vr=bk$, then proving $b\geq v$ is the same as proving $r\geq k$. So in your case it is enough to show that $r\geq 4$, i.e. that each element is contained in 4 or more blocks. Moreover, the second basic equality $r(k-1)=\lambda(v-1)$ in common with $v>k$ imply that $r>\lambda$. Now it is enough to play with cases $(r,\lambda)\in\{(2,1),(3,1),(3,2)\}$. However, again, using the basic two equalities and $k=4$ we obtain that $$ b = \frac r4\cdot\left(\frac{3r}{\lambda}+1\right).$$ None of the 3 cases lead to integer $b$, therefore necessarily $r\geq 4$, and we are ready.