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The answer to this question is yes.

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$, and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. 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} and then 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 $u_+^a:=\max(0,u)^a$ and $\mu=\mu_Y$ is the probability distribution of $Y$, with the random variable (r.v.) $X$ considered fixed.

Obviously, the function $F$ is convex on the simplex $M_S$ of all probability distributions $\mu_Y$ on $S$. So, the maximum of $F$ on $M_S$ is attained when $\mu$ is a Dirac measure -- that is, when the r.v. $Y$ is a nonnegative constant $c$, so that
$F(\mu)=c^a(EX^{a-1}-E^{a-1}X)_+^a=0$, because $1<a\le2$ and hence $EX^{a-1}\le E^{a-1}X$. So, the maximum of $F$ on the simplex $M_S$ is $0$, which completes the proof for the case $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.

The answer to this question is yes.

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$, and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. 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 $\frac d{dt}h_{X,Y}(t)=\frac d{du}h_{X+tY,Y}(u)|_{u=0}$). 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} and then 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 $u_+^a:=\max(0,u)^a$ and $\mu=\mu_Y$ is the probability distribution of $Y$, with the random variable (r.v.) $X$ considered fixed.

Obviously, the function $F$ is convex on the simplex $M_S$ of all probability distributions $\mu_Y$ on $S$. So, the maximum of $F$ on $M_S$ is attained when $\mu$ is a Dirac measure -- that is, when the r.v. $Y$ is a nonnegative constant $c$, so that
$F(\mu)=c^a(EX^{a-1}-E^{a-1}X)_+^a=0$, because $1<a\le2$ and hence $EX^{a-1}\le E^{a-1}X$. So, the maximum of $F$ on the simplex $M_S$ is $0$, which completes the proof for the case $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.

The answer to this question is yes.

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$ and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. 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} and then 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 $u_+^a:=\max(0,u)^a$ and $\mu=\mu_Y$ is the probability distribution of $Y$, with the random variable (r.v.) $X$ considered fixed.

Obviously, the function $F$ is convex on the simplex $M_S$ of all probability distributions $\mu_Y$ on $S$. So, the maximum of $F$ on $M_S$ is attained when $\mu$ is a Dirac measure -- that is, when the r.v. $Y$ is a nonnegative constant $c$, so that
$F(\mu)=c^a(EX^{a-1}-E^{a-1}X)_+^a=0$, because $1<a\le2$ and hence $EX^{a-1}\le E^{a-1}X$. So, the maximum of $F$ on the simplex $M_S$ is $0$, which completes the proof for the case $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.

The answer to this question is yes.

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$, and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. 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} and then 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 $u_+^a:=\max(0,u)^a$ and $\mu=\mu_Y$ is the probability distribution of $Y$, with the random variable (r.v.) $X$ considered fixed.

Obviously, the function $F$ is convex on the simplex $M_S$ of all probability distributions $\mu_Y$ on $S$. So, the maximum of $F$ on $M_S$ is attained when $\mu$ is a Dirac measure -- that is, when the r.v. $Y$ is a nonnegative constant $c$, so that
$F(\mu)=c^a(EX^{a-1}-E^{a-1}X)_+^a=0$, because $1<a\le2$ and hence $EX^{a-1}\le E^{a-1}X$. So, the maximum of $F$ on the simplex $M_S$ is $0$, which completes the proof for the case $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.

The answer to this question is yes.

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$ and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. 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 set $M$ 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 $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.

The answer to this question is yes.

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$ and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. 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} and then 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 $u_+^a:=\max(0,u)^a$ and $\mu=\mu_Y$ is the probability distribution of $Y$, with the random variable (r.v.) $X$ considered fixed. 

Obviously, the function $F$ is convex on the simplex $M_S$ of all probability distributions $\mu_Y$ on $S$. So, the maximum of $F$ on $M_S$ is attained when $\mu$ is a Dirac measure -- that is, when the r.v. $Y$ is a nonnegative constant $c$, so that
$F(\mu)=c^a(EX^{a-1}-E^{a-1}X)_+^a=0$, because $1<a\le2$ and hence $EX^{a-1}\le E^{a-1}X$. So, the maximum of $F$ on the simplex $M_S$ is $0$, which completes the proof for the case $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.