Consider a slowly varying function $L(n)$ and let $\lambda > 0$. Is it true that there exists a $N \in \mathbb{N}$ such that
\begin{align} \frac{L(\lambda n)}{L(n)} \geq 1 \end{align} for all $n \geq N$? And, if not, are there any conditions under which this holds true?
It is not true: consider $L(x) = \log(x)$, then for $n\geq 3$ we have that
$$ \dfrac{L(\lambda n)}{L(n)} = \dfrac{\log(\lambda)}{\log(n)} + 1 < 1 $$
if $\lambda\in(0,1)$. I am not aware of conditions in order for what you want to be true.