= Solution
Define the <barycenter of a measure> by $m_\mu=\int x\,d\mu(x)$. Under $\Pi$, the mean displacement is $\bar q=m_\nu-m_\mu$, and optimality gives $\int|q|^2\,d\Pi=D^2$. Expanding the square under the <product measure> yields the <variance identity for independent copies>:
$$
\frac12\iint|q-q'|^2\,d\Pi\,d\Pi'=\int|q|^2\,d\Pi-\left|\int q\,d\Pi\right|^2=D^2-|m_\nu-m_\mu|^2.
$$
In the classical regime of part (c), the lower bound on the <Hessian matrix> of the <interaction potential> therefore gives
$$
\frac{d^2}{dt^2}\mathcal W[\mu_t]\geq\lambda\left(D^2-|m_\nu-m_\mu|^2\right).
$$
For equal <barycenters>, this becomes
$$
\boxed{\frac{d^2}{dt^2}\mathcal W[\mu_t]\geq\lambda D^2.}
$$
For the full stated class of <probability measures>, prove the <displacement convexity> inequality directly. Pointwise strong <convexity> of $W$ gives
$$
W((1-t)(x-x')+t(y-y'))\leq(1-t)W(x-x')+tW(y-y')-\frac\lambda2t(1-t)|q-q'|^2.
$$
Integrating, including the $1/2$ in the <interaction energy>, and using the displayed <variance identity for independent copies> gives
$$
\mathcal W[\mu_t]\leq(1-t)\mathcal W[\mu]+t\mathcal W[\nu]-\frac\lambda2t(1-t)\left(D^2-|m_\nu-m_\mu|^2\right).
$$
All terms are finite by quadratic growth and finite second <absolute moments>. Equal <barycenters> give the claimed $\lambda$-<displacement convexity>. Also $m_{\mu_t}=(1-t)m_\mu+tm_\nu$, so this <metric geodesic> remains in the prescribed fixed-mean class. Applying the same argument to subintervals proves the second derivative bound in the sense of <distributions> even when a classical derivative has not been justified. The mean restriction is natural: translating all particles together leaves their pairwise differences and hence the <interaction energy> unchanged.
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