= Solution
In the <classical risk model>, the surplus with initial capital $u\geq0$ is
$$
U(t)=u+ct-\sum_{j=1}^{N(t)}X_j,\qquad c=(1+\rho)\lambda\mu,
$$
where $N(t)$ is a <Poisson process> of rate $\lambda$, and the nonnegative independent <claim sizes>, of mean $\mu>0$, are independent of that process. Positive <relative safety loading> $\rho$ makes expected premium income exceed expected claim outflow. With ruin time $T=\inf\{t\geq0:U(t)<0\}$, the <ultimate ruin probability> is
$$
\boxed{\psi(u)=\mathbb P(T<\infty\mid U(0)=u).}
$$
The <adjustment coefficient> is the positive root $R$ in the finite domain of the claim <moment-generating function> satisfying
$$
\boxed{\lambda(M_X(R)-1)=cR,\quad\text{or }M_X(R)=1+(1+\rho)\mu R.}
$$
The zero root is excluded by definition.
Under the stated transform-divergence assumption this positive root exists and is unique. Indeed $F(r)=M_X(r)-1-(1+\rho)\mu r$ has $F(0)=0$, derivative $F'(0)=-\rho\mu<0$, and strictly positive second derivative on its positive domain. At a finite endpoint the diverging transform makes $F$ positive eventually. At an infinite endpoint, choose $a>0$ with $\mathbb P(X\geq a)>0$; the bound $M_X(r)\geq\mathbb P(X\geq a)e^{ar}$ makes it dominate the linear term. Strict convexity then gives exactly one positive crossing. This is the <secant-slope existence criterion for an adjustment coefficient>.
The <Lundberg inequality> is
$$
\boxed{\psi(u)\leq e^{-Ru}\qquad(u\geq0).}
$$
For a short justification, the compound-Poisson transform and the defining equation make $e^{-RU(t)}$ a nonnegative <martingale> starting at $e^{-Ru}$. Stop it at $T\wedge t$, a bounded <stopping time>. On $\{T\leq t\}$ its value is greater than one. The <optional stopping theorem> therefore gives $\mathbb P(T\leq t)\leq e^{-Ru}$; let $t$ increase to infinity. No expectation identity at an unbounded ruin time is required.
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