Suppose $T = \inf\{t: B_t\not\in (a,b)$ where $a<0<b$ and $a\neq -b$. I would like to show $$\mathbb{E}T^2 \leq C \mathbb{E}B_T^4.$$
The problem also says to apply Cauchy-Schwarz inequality to $\mathbb{E}(TB_T^2)$.
Now I know $\{B_t^4 - 6tB_t^2 + 3t^2\}_t$ is a martingale, and it suffices to show the inequality above for $T\wedge t$. By martingale property and Cauchy-Schwarz, we have $$\mathbb{E} B_{T\wedge t}^4 + 3\mathbb{E} (T\wedge t)^2 \leq 6 \bigg(\mathbb{E} (T\wedge t)^2\bigg)^{1/2} \bigg( \mathbb{E} B_{T\wedge t}^4\bigg)^{1/2}$$ I am stuck here, and I am trying to conclude from this without getting into too explicit calculation with the $B_T$ term since we can actually compute $\mathbb{E}T^2$ and $\mathbb{E}B_T^4$, but I think it is not the point of this problem...
It is just a scaling argument now. Write $E[T^2]=C(a,b) E[B_T^4]$ and then your inequality becomes
$$(3C(a,b)+1)E[B_T^4] \leq 6 C(a,b)^{1/2} E[B_T^4].$$
Thus $3C(a,b)+1 \leq 6C(a,b)^{1/2}$. There is a maximal solution to this inequality.
The point is that $E[T^2]$ contributes more to the left side of the inequality than to the right (the exponent is $1$ vs. $1/2$), so unless the factor $E[B_T^4]$ can "pick up the slack" on the RHS, the inequality you know to be true cannot be true. You could use exactly the same argument to derive the reverse inequality (for a different $C$).