I need to show that if $(X_j)$ are symmetric i.i.d. random variables with partial sums $S_n:= \sum_{j=1}^n X_j$, then for all $x \geq 0$
$$P(|S_n| > x) \geq \frac{1}{2} P(\max_{1 \leq j \leq n} |X_j| > x). $$
My attempt is to define an event $$\Delta:=\{\max_{1 \leq j \leq n} |X_j| > x\}$$ and to define on $\Delta$ the random variable $\tau$ to be the first $j$ such that $|X_j| > x$ (and put $\tau = \infty$ on $\Delta^c$). Fixing $n$ and writing $$S_j':= \sum_{1 \leq i \leq n, i \neq j } X_i \quad ,$$
I can write
$$P(|S_n| > x) \geq \sum_{j=1}^n P( \mathrm{sign}(S_j')=\mathrm{sign}(X_j); \tau = j ) .$$
The event $A:= \{ \mathrm{sign}(S_j')=\mathrm{sign}(X_j) \}$ has probability $1/2$ by symmetry and independence, so we could conclude if only $A$ and $\{\tau= j\}$ were independent. Although this seems plausible, I'm not sure it's true, or how to prove it.
Resolution: By conditioning on the variables involved in the event $A \cap \{\tau=j\}$, it is possible to switch the signs of the $X_k \hspace{3pt} (k \neq j)$ to prove the unsurprising equation:
$$ P( \mathrm{sign}(S_j')=\mathrm{sign}(X_j); \tau = j ) = P( -\mathrm{sign}(S_j')=\mathrm{sign}(X_j); \tau = j ) ,$$
which is enough to conclude. The key is to use condition to swap the signs one by one.