Context:
A solution $(M^n, g(t))$ of the Ricci flow is said to encounter a Type III Singularity if $g(t)$ is defined for all $t \geq 0$ and: $$ \sup _{\mathcal{M}^{n} \times[0, \infty)} \|\operatorname{Rm}(\cdot, t) \| t<\infty $$ Similarly, if $g(t)$ is defined for all $t \geq 0$ but: $$ \sup _{\mathcal{M}^{n} \times[0, \infty)} \|\operatorname{Rm}(\cdot, t) \| t= \infty $$ the solution is said to encounter a Type IIb singularity.
Let $(M, g_0)$ be an Einstein manifold with $\text{Ric}_{g_0} = \lambda_0 g_0$. Then a solution to its Ricci flow is the family of metrics given by $g(t) = (1- 2\lambda_0 t)g_0$, which has scalar curvature given by $R(t) = \frac{\lambda_0 n}{1-2t}$. I'm trying to prove that if $\lambda_0 < 0$, a type III singularity happens and if $\lambda_0 = 0$, a type IIb singularity happens. In those cases obviously $g(t)$ is defined for all $t \geq 0$, but assuming without loss of generality that $\lambda_0 = -\frac{1}{2}$ in the first case, we have:
$$\|\text{Rm}(x, t)\| t = \frac{t}{1+t} \|\text{Rm}(x,0) \| \ \forall (x, t) \in M \times [0, \infty)$$
But I don't understand how we can bound this. And in the second case $\lambda_0 = 0$ we get$$\|\text{Rm}(x, t)\|t = t \|\text{Rm}(x,0) \|$$
But again, how is this unbounded? We only have control on the Ricci tensor, not on the curvature tensor (unless we're in dimension $\leq 3$ but that's a silly hypothesis).
EDIT: The case $\lambda_0 = 0$ is actually trivial, since for each $x_0 \in M$, it's clear that $\{\|\text{Rm}(x_0, t)\|t \ \vert \ t \geq 0 \}$ is unbounded.