The inequality is false in general. E.g., let $\alpha = 5777/2500$, $X:=f(I,J)$, and $Y:=g(I,J)$, where $I$ and $J$ are random variables with values in the sets $\{1,2,3\}$ and $\{1,2\}$, respectively, such that $P(I=i,J=j)=p(i,j)$ for all $i\in\{1,2,3\}$ and $j\in\{1,2\}$, where the matrices $f=(f(i,j))$, $g=(g(i,j))$, and $p=(p(i,j))$ are as follows: \begin{align*} f=\frac1{10^5} \left( \begin{array}{cc} 11853 & 13956 \\ 13544 & 2908 \\ 11609 & 9967 \\ \end{array} \right),\quad g=\frac1{10^5} \left( \begin{array}{cc} 10992 & 13254 \\ 14883 & 8078 \\ 11782 & 11134 \\ \end{array} \right),\quad \end{align*} \begin{equation*} p=\frac1{10^5} \left( \begin{array}{cc} 297 & 535 \\ 2194 & 6244 \\ 356 & 374 \\ \end{array} \right). \end{equation*}
The the difference between the LHS and RHS of your inequality is $-0.00696\ldots<0$.
Let now $a:=\alpha$, and suppose $1<a\le2$. Without loss of generality (wlog), $EX^a<\infty$ and $Y$ takes only values in a fixed finite set. The inequality in question is equivalent to the following: \begin{equation} h(t):=h_{X,Y}(t):=[E(X+tY)^a-E^a(X+tY)]^{1/a}-[(EX^a-E^a X)^{1/a}+t(EY^a-E^a Y)^{1/a}]\le0 \end{equation} for all $t\ge0$, and hence to the condition $h'(0)\le0$ (because $\frac d{dt}h_{X,Y}(t)=\frac d{du}h_{X+(t+u)Y,Y}(u)|_{u=0}$. In turn, the condition $h'(0)\le0$ can be rewritten as \begin{equation} (EX^{a-1}Y-E^{a-1}X\,EY)(EX^a-E^a X)^{1/a-1}\le(EY^a-E^a Y)^{1/a}. \end{equation} The latter inequality is trivial if $EX^{a-1}Y-E^{a-1}X\,EY\le0$. Otherwise, it can be rewritten in this H"older-like form: \begin{equation} F(\mu):=(EX^{a-1}Y-E^{a-1}X\,EY)^a-(EX^a-E^a X)^{a-1}(EY^a-E^a Y)\le0, \end{equation} where $\mu$ is the probability distribution of $Y$. Obviously, the function $F$ is convex. So, its maximum on the simplex of all possible $\mu$'s is attained when $\mu$ is a Dirac measure, that is, when $Y$ is a constant random variable; then wlog $Y=1$, so that $F(\mu)=EX^{a-1}-E^{a-1}X\le0$, since $1<a\le2$.