$X_1$ and $X_2$ stochastically larger than $Y_1$ and $Y_2$ implies $X_1+X_2$ stochastically larger than $Y_1+Y_2$ The random variables $X_1$, $X_2$, $Y_1$ and $Y_2$ are all mutually independent and I think one can show this in two different ways. In one paper I just came to a point where this may need to be shown. But I am not sure whether it is trivial or not. 
The simplest idea is probably to show that  $$G_{Y_1+Y_2}(x)\geq F_{X_1+X_2}(x)\quad \forall x$$  where $G_{Y_1+Y_2}$ and $F_{X_1+X_2}$ are the distribution functions of $Y_1+Y_2$ and $X_1+X_2$ respectively. From stochastic ordering it is also known that $G_{Y_1}(x)\geq F_{X_1}(x)$ and $G_{Y_2}(x)\geq F_{X_2}(x)$ for all $x$. 

Do you think that the question is trivial? could one show it?


\begin{align}
G_{Y_1+Y_2}(z)&=\int_{-\infty}^{+\infty}G_{Y_1}(z-x)dG_{Y_2}(x)
\ge \int_{-\infty}^{+\infty}F_{X_1}(z-x)dG_{Y_2}(x)\\
&=\iint_{x+y\le z}dF_{X_1}(x)dG_{Y_2}(y)
= \int_{-\infty}^{+\infty}G_{Y_2}(z-y)dF_{X_1}(y)\\
&\ge \int_{-\infty}^{+\infty}F_{X_2}(z-y)dF_{X_1}(y)=F_{X_1+X_2}(z).
\end{align}
