= Solution
Let $F$ be the distribution function of $\xi_1$, write $\overline F=1-F$, and fix $x>0$. Set
$$
q(t)=\mathbb P(L(t)\geq x).
$$
Conditioning on the first interarrival gives
$$
q(t)=\overline F(\max\{x,t\})
+\int_{[0,t]}q(t-y)\,dF(y).
$$
Indeed, if $\xi_1>t$, the first interval contains $t$ and must also have length at least $x$; if $\xi_1\leq t$, the process starts afresh at time $\xi_1$.
Put $h=\overline F(x)$ and $d(t)=q(t)-h$. Then
$$
d(t)=g(t)+\int_{[0,t]}d(t-y)\,dF(y),
$$
where
$$
g(t)=
\begin{cases}
hF(t),&t\leq x,\\
\overline F(t)F(x),&t>x.
\end{cases}
$$
In particular $g\geq0$. Iterating the renewal equation yields
$$
d(t)=\sum_{n=0}^{m}(g*dF^{*n})(t)
+(d*dF^{*(m+1)})(t).
$$
For fixed $t$, the last term tends to zero in absolute value because $d$ is bounded and
$$
F^{*(m+1)}(t)=\mathbb P(S_{m+1}\leq t)\longrightarrow0.
$$
It follows that
$$
d(t)=\sum_{n=0}^{\infty}(g*dF^{*n})(t)\geq0.
$$
Thus the <stochastic length bias of a renewal interval> gives
$$
\boxed{\mathbb P(L(t)\geq x)\geq\mathbb P(\xi_1\geq x)}
$$
for every $x,t>0$.
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