On a filtered probability space, supporting a standard Brownian motion $(W_t)_{t\geq 0}$ and an adapted $\textbf{non-decreasing}$, integrable, continuous process $(A_t)_{t\geq 0}$ starting at $A_0=0$ (so a finite variation process). Consider the following two processes $Y$ and $Z$ with $Z_0=Y_0=x$ such that $x>0$, $$Y_t=x+\mu\int_0^t\mathcal{E}(\sigma W_s)^{-1}e^{-A_s}dA_s$$ and $$Z_t=x+\mu\mathcal{E}(\sigma W_t)^{-1}(1-e^{-A_t})$$ where $\mu,\sigma>0$ and $\mathcal{E}$ denotes the stochastic exponential, so $\mathcal{E}(\sigma W_s)=e^{\sigma W_t-\frac{\sigma^2t}{2}}$.
Question: Let $0<\gamma<1$. Show that $$\mathbb{E}\Big[e^{-\gamma A_t}\mathcal{E}(\sigma W_t)^{1-\gamma}Y_t^{1-\gamma}\Big]\leq \mathbb{E}\Big[e^{-\gamma A_t}\mathcal{E}(\sigma W_t)^{1-\gamma}Z_t^{1-\gamma}\Big]$$ for all $t\geq 0$.
My attempt: My thought is to use Ito formula and take expectation, then get rid of the martingale parts and compare the drift.
Let $Q_t:=e^{-\gamma A_t}\mathcal{E}(\sigma W_t)^{1-\gamma}$, so the above inequaility beomes $$\mathbb{E}\Big[Q_tY_t^{1-\gamma}\Big]\leq \mathbb{E}\Big[Q_tZ_t^{1-\gamma}\Big].$$
By Ito formula, $$dZ_t=\mu\mathcal{E}(\sigma W_t)^{-1}\Big( e^{-A_t}dA_t+\sigma^2(1-e^{-A_t})dt-\sigma(1-e^{-A_t})dW_t\Big)$$ and $$\frac{dQ_t}{Q_t}=-\gamma dA_t-\gamma(1-\gamma)\sigma^2dt+\sigma(1-\gamma) dW_t.$$ Then $$d(Q_tY_t^{1-\gamma})=Y_t^{-\gamma}Q_t\Big[\Big(\mu(1-\gamma)\mathcal{E}(\sigma W_t)^{-1}e^{-A_t}-\gamma Y_t\Big)dA_t-\gamma(1-\gamma)\sigma^2Y_tdt\Big]+dM_t$$ and $$d(Q_tZ_t^{1-\gamma})=(1-\gamma)\sigma^2Z_t^{-\gamma}Q_t\Big(\frac{\mu(1+\gamma)}{2}\mathcal{E}(\sigma W_t)^{-1}(1-e^{-A_t})-\gamma Z_t-\frac{\gamma\mu^2}{2}Z_t^{-1}\mathcal{E}(\sigma W_t)^{-2}(1-e^{-A_t})^2\Big)dt+Z_t^{-\gamma}Q_t\Big(\mu(1-\gamma)\mathcal{E}(\sigma W_t)^{-1}e^{-A_t}-\gamma Z_t\Big)dA_t+dM_t$$ where $M$ denotes some martingale. To replace the term with $Z$, we obtain $$d(Q_tZ_t^{1-\gamma})=(1-\gamma)\sigma^2Z_t^{-\gamma}Q_t\Big(\frac{(1+\gamma)}{2}(Z_t-x)-\gamma Z_t-\frac{\gamma}{2}Z_t^{-1}(Z_t-x)^2\Big)dt+Z_t^{-\gamma}Q_t\Big(\mu(1-\gamma)\mathcal{E}(\sigma W_t)^{-1}e^{-A_t}-\gamma Z_t\Big)dA_t+dM_t.$$
But then when I take expectation, the integral over $dA_t$ remains (i.e. $\mathbb{E}[\int_0^t\cdots dA_t]=?$) as I cannot take the expectation inside the integral, so I have no clue how to proceed.