= Solution
Write $V_t=\langle X\rangle_t$ for the <quadratic variation> and set
$$
C(p,q)=\frac{pq-\sqrt{pq}}{q-1}
=\frac{\sqrt{pq}(\sqrt{pq}-1)}{q-1}.
$$
The <Hölder factorization of stochastic exponentials> follows by adding exponents:
$$
\begin{aligned}
\frac1q\left(\sqrt{pq}X_t-\frac{pq}{2}V_t\right)
+\frac{q-1}{q}C(p,q)X_t
&=\frac{\sqrt{pq}+pq-\sqrt{pq}}qX_t-\frac p2V_t\\
&=pX_t-\frac p2V_t.
\end{aligned}
$$
Thus
$$
\boxed{\mathcal E(X)_t^p
=\mathcal E(\sqrt{pq}X)_t^{1/q}
\bigl(e^{C(p,q)X_t}\bigr)^{(q-1)/q}.}
$$
The subtraction inside the numerator is $\sqrt{pq}-1$, outside the square root.
The <stochastic exponential> $\mathcal E(\sqrt{pq}X)$ starts at one and is a <nonnegative local martingale>, hence a <supermartingale>. The <optional sampling theorem for a supermartingale> gives expectation at most one at bounded stopping times. For a finite, possibly unbounded, stopping time $T$, apply this to $T\wedge n$, then use continuity and the <Fatou lemma>:
$$
\mathbb E\mathcal E(\sqrt{pq}X)_T\leq1.
$$
The <Holder inequality> with exponents $q$ and $q/(q-1)$ now gives
$$
\boxed{
\mathbb E[\mathcal E(X)_T^p]
\leq
\bigl(\mathbb E[e^{C(p,q)X_T}]\bigr)^{(q-1)/q}.}
$$
The inequality also holds when the right side is infinite.
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