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 $S\subset \mathbb R$. 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 $\mu=\mu_Y$ is the probability distribution of $Y$, with random variable (r.v.) $X$ considered fixed. Obviously, the function $F$ is convex on the $M$ of all probability distributions $\mu_Y$ on $S$ such that $EX^{a-1}Y-E^{a-1}X\,EY\ge0$. So, the maximum of $F$ on the compact convex set $M$ is attained when either $\mu$ is a Dirac measure (that is, when the r.v. $Y$ is a nonnegative constant) or at a boundary point of $M$ where $EX^{a-1}Y-E^{a-1}X\,EY=0$. 
However, if $Y$ is a nonnegative constant $c$, then $EX^{a-1}Y-E^{a-1}X\,EY=c(EX^{a-1}-E^{a-1}X)\le0$, since $1<a\le2$. So, the maximum of $F$ on the compact convex set $M$ is attained when $EX^{a-1}Y-E^{a-1}X\,EY=0$, and then clearly $F(\mu)\le0$. 
This completes the proof for the case $1\le a\le2$.