= Solution
The severity is an equal-weight <mixture distribution> of rate-$2$ and rate-$2/3$ <exponential distributions>. Its <expected value> is
$$
\boxed{\mu=\frac12\frac12+\frac12\frac32=1.}
$$
The <moment-generating function>, finite exactly for $r<2/3$, is
$$
M_X(r)=\frac1{2-r}+\frac1{2-3r},\qquad
M_X(r)-1=\frac{r(4-3r)}{(2-r)(2-3r)}.
$$
Canceling the nonzero $r$ in the <adjustment coefficient> equation and multiplying by the positive denominator on $0<r<2/3$ gives
$$
4-3r=(1+\theta)(4-8r+3r^2).
$$
Thus the required <polynomial> is
$$
P(r)=3(1+\theta)r^2-(8\theta+5)r+4\theta.
$$
Since $P(0)=4\theta>0$, $P(2/3)=-2<0$, and the leading coefficient is positive, one root lies in $(0,2/3)$ and the other exceeds $2/3$. Only the smaller root is in the <moment-generating function> domain. Consequently
$$
\boxed{R_{\rm mix}=\frac{8\theta+5-\sqrt{16\theta^2+32\theta+25}}{6(1+\theta)}
=\frac{8\theta}{8\theta+5+\sqrt{16\theta^2+32\theta+25}}.}
$$
The rationalized expression avoids subtraction of nearly equal numbers when $\theta$ is small.
The erroneously chosen same-mean <exponential distribution> gives $R_{\rm exp}=\theta/(1+\theta)$. Direct substitution shows $P(R_{\rm exp})=-\theta<0$, so $R_{\rm exp}$ lies strictly between the two polynomial roots, and in particular $R_{\rm mix}<R_{\rm exp}$. Therefore
$$
\boxed{e^{-R_{\rm exp}u}<e^{-R_{\rm mix}u}\quad(u>0),}
$$
while both bounds equal one at $u=0$. \b[The exponential misspecification produces a smaller claimed upper bound.] The <Lundberg inequality> certifies $e^{-R_{\rm mix}u}$ for the actual mixture; it does not certify the smaller expression computed from the wrong severity law.
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