= Solution
In the <classical risk model>, write the surplus as
$$
U(t)=u+ct-\sum_{j=1}^{N(t)}X_j,\qquad c=(1+\theta)\lambda\mu,
$$
where $N(t)$ is the claim-arrival <Poisson process>. The ruin time is $\tau=\inf\{t\ge0:U(t)<0\}$. The <ultimate ruin probability> is
$$
\boxed{\psi(u)=\mathbb P(\tau<\infty).}
$$
The positive root $R$ in the question is the <adjustment coefficient>. The <Lundberg inequality> and the <Cramér–Lundberg approximation> are, respectively,
$$
\boxed{\psi(u)\le e^{-Ru},}
\qquad
\boxed{\psi(u)\sim C e^{-Ru}\quad(u\to\infty),\qquad
C=\frac{\theta\mu}{M_X'(R)-(1+\theta)\mu}.}
$$
The <interior adjustment coefficient ruin prefactor> is finite here because $R$ lies strictly inside the finite domain of the <moment-generating function>; the claim-size density supplies the nonarithmetic condition for the asymptotic. Strict convexity makes its denominator positive. The approximation is an asymptotic statement, while the inequality is a bound at every nonnegative initial capital.
The inequality can also be seen directly from the <exponential surplus martingale> $e^{-RU(t)}$. Its expectation factor per unit time is zero in the exponent because $\lambda(M_X(R)-1)-cR=0$. Apply the <optional stopping theorem> at $\tau\wedge T$, a bounded time. On $\{\tau\le T\}$, $e^{-RU(\tau)}>1$, so $\mathbb P(\tau\le T)\le e^{-Ru}$. Increasing $T$ proves the bound without stopping at an unbounded time.
For any nonnegative claim size, the <Tonelli theorem> applied to $e^{RX}-1=\int_0^X Re^{Rx}\,dx$ gives the exponential tail identity
$$
M_X(R)-1=R\int_0^\infty e^{Rx}(1-F_X(x))\,dx.
$$
Combining it with the adjustment equation yields
$$
\boxed{\frac1\mu\int_0^\infty e^{Rx}(1-F_X(x))\,dx=1+\theta.}
$$
The density $f_Y(x)=(1-F_X(x))/\mu$ integrates to one by the <tail integral formula for moments>. Thus $Y$ has the <equilibrium claim-size distribution>, and the preceding formula says $M_Y(R)=1+\theta$.
Under the given <NBUE> condition, the <stochastic order> comparison is $1-F_Y(x)\le1-F_X(x)$ at every $x\ge0$. Apply the same exponential tail identity to both variables to obtain
$$
\mathbb E e^{RY}=1+R\int_0^\infty e^{Rx}(1-F_Y(x))\,dx
\le1+R\int_0^\infty e^{Rx}(1-F_X(x))\,dx
=\mathbb E e^{RX}.
$$
Consequently
$$
1+\theta=M_Y(R)\le M_X(R)=1+(1+\theta)\mu R,
$$
and rearrangement proves the <adjustment-coefficient lower bound for NBUE claims>:
$$
\boxed{R\ge\frac{\theta}{(1+\theta)\mu}.}
$$
This uses the specified distributional comparison, not an unwarranted assumption that an arbitrary equilibrium claim law is always smaller than its original law.
For the final example, $f_X(x)=xe^{-x}$ is the shape-two, rate-one <gamma distribution>, with $\mu=2$, survival function $\overline F_X(x)=(1+x)e^{-x}$, and <moment-generating function> $M_X(r)=(1-r)^{-2}$ for $r<1$. With $\theta=1/10$, the positive adjustment root satisfies
$$
(1-R)^{-2}-1=\frac{11}{5}R.
$$
Dividing out the zero root and simplifying gives $11R^2-17R+1=0$. Only the smaller root is in the finite-transform interval $(0,1)$, so
$$
\boxed{R=\frac{17-7\sqrt5}{22}\approx0.0612511.}
$$
The equilibrium density and survival function are
$$
f_Y(x)=\frac{(1+x)e^{-x}}2,\qquad
\overline F_Y(x)=\frac12\int_x^\infty(1+t)e^{-t}\,dt
=(1+x/2)e^{-x}.
$$
Thus $\overline F_Y(x)\le(1+x)e^{-x}=\overline F_X(x)$ for every $x\ge0$, proving that this claim law is <new better than used in expectation>. Finally the lower bound equals $1/22$. Our exact root is larger because $16>7\sqrt5$, equivalently $256>245$. Hence
$$
\boxed{R>\frac1{22}\approx0.0454545,}
$$
verifying the requested bound for the example.
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