In the accepted answer in this post it is claimed that Brownian motions have the property that $$\sup\limits_{s \leq t} \frac{B_s}{\sqrt{s}} \leq \frac{B_t}{\sqrt{t}}$$
Can someone help me understand why is this true? I assume this really means that for any $C \in \mathbb{R}$, $s \in (0,t]$: $$P\bigg(\frac{B_s}{\sqrt{s}} < C \bigg) \leq P\bigg(\frac{B_t}{\sqrt{t}} < C \bigg)$$
If true then this would be analogous to prove that for all $C,s$ as before:
$$1 \leq \sqrt{\frac{s}{t}}\frac{\int_{-\infty}^{C\sqrt{t}}e^{x^2/2t}dx}{\int_{-\infty}^{C\sqrt{s}}e^{y^2/2t}dy}$$
But again, why does this inequality hold?
As @Kurt mentioned in his comment, the inequality should be flipped: $$\sup_{s \leq t} \frac{B_s}{\sqrt{s}} \geq \frac{B_t}{\sqrt{t}}$$ where this inequality follows since the supremum is taken over a set which includes $\frac{B_t}{\sqrt{t}}$.
There is, however, a misreading of the linked post, which says that "$\sup_{s \leq t} \frac{B_s}{\sqrt{s}}$ is monotonic with respect to $t$". This means that if you define the process $M_t = \sup_{s \leq t} \frac{B_s}{\sqrt{s}}$, then the process is non-decreasing; that is , for $r \leq t$ we have $M_r \leq M_t$. Again, this follows as for $t$ the supremum is taken over a larger set.