Let $X$ be a random variable such that $E[X]<+\infty$, and $Y$ be a random variable independent of $X$ such that $E[Y]=0$. Define $Z=X+Y$. Let $g:\mathbb{R}\to\mathbb{R}$ be a concave function.
Show that $E[g(X)]\geq E[g(Z)]$.
I don't see how to use the concavity of $g$ since $Z$ is not a convex combination of $X$ and $Y$. I don't think Jensen's inequality can be used here...
You may require the conditional expectation version of jensen's inequality
$$E[g(Z)]=E[g(X+Y)]= E[E[g(X+Y)|X]] \leq E[g(E[X+Y|X])]$$ $$= E[g(X+E[Y|X])] = E[g(X)]$$ where we used that $E[Y|X]=E[Y]=0$ as $X$ and $Y$ are independent.
Suppose, just to fix ideas, that both $X$ and $Y$ are continuous r.v.s that admit pdf. Suppose $g$ is concave. Then $$\mathbb{E}[g(X+Y)]=\int\int g(x+y)f_{Y}(y)f_{X}(x)dydx\leq\int g\left(\int(x+y)f_{Y}(y)dy\right)f_{X}(x)dx=\int g(x)f_{X}(x)dx=\mathbb{E}[g(X)]$$ where the inequality follows by Jensen's inequality.
As a side remark, in economics (choice under uncertainty), $Y$ is called mean-preserving spread of $X$. One can define '$X$ second-order stochastically dominates $Y$' if expected utility of risk averse decision maker is weakly higher under $X$ than under $Y$. Then '$X$ is mean-preserving spread of $Y$' and '$X$ second-order stochastically dominates $Y$ are equivalent'. See Section 6.D in this book.