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For $t>0$, put $M_t={\sup }_{0\le s\le t}{B}_s$. For $\lambda >0$, reflect the Brownian path after its first hit of $\lambda$. The Strong Markov property and symmetry of subsequent Brownian increments make this reflection preserve the path distribution. It exchanges the events $\{M_t\ge \lambda ,B_t<\lambda \}$ and $\{B_t>\lambda \}$. Since $B_t$ has no atom at $\lambda$, the Brownian reflection principle givesThe case $\lambda =0$ follows by decreasing positive levels to zero: the probability is one. At $t=0$, $M_0=0$ and the probability is zero for every $\lambda \ge 0$.

The displayed tail probabilities are exactly those of $|B_t|$, so $M_t$ has the half-normal distribution of $\sqrt{t}|Z|$ with $Z\sim N(0,1)$. ThusFor $A_t={\sup }_{s\le t}|B_s|$, the pointwise inequality $A_t\ge M_t$ gives the lower non-strict bound. To prove strictness when $t>0$, fix $\lambda >0$. Reflection at $\lambda$ also givesSubtracting this from $\mathbb{P}(B_t<-\lambda )=\mathbb{P}(B_t>\lambda )$ givesOn this event $A_t\ge |B_t|>\lambda >M_t$, so $A_t^2-M_t^2$ is strictly positive with positive probability. Finally, $|B_t|$ is a continuous nonnegative submartingale by the conditional Jensen inequality, and part (a) gives $\Vert A_t{\Vert }_2\le 2\Vert B_t{\Vert }_2=2\sqrt{t}$. Its square is therefore integrable, and the strict pointwise comparison on a positive-probability event givesThis is the strict comparison of one-sided and absolute Brownian maxima. The strict lower inequality printed with $t\ge 0$ needs the qualification $t>0$: at $t=0$ both maxima vanish and all three quantities are zero.