Proof of Inequality (6)

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Let $X$ and $Y$ be random variables.  If $X\geq0$ and $Y\geq0$, $p\geq0$, then $$\mathscr{E}\{(X+Y)^p\}\leq2^p\{\mathscr{E}(X^p)+\mathscr{E}(Y^p)\}.$$If $p>1$, the factor $2^p$ may be replaced by $2^{p-1}$. If $0\leq p\leq1$, it may be replaced by $1$.

See List of Inequalities.

$\bullet$ Proof.

Define $\phi(x)=x^p$, when $x\geq0$, which is convex if $p>1$,
$i.e.\,\forall\,x,y\in[0,\infty),$ $$\phi(\alpha x+(1-\alpha)y)\leq\phi(\alpha x)+\phi((1-\alpha)y).$$WLOG, let $\alpha = 1/2$, we have $$\begin{array}{rl}
            &\, \left(\frac{1}{2}x+\frac{1}{2}y\right)^p\leq\frac{1}{2}x^p+\frac{1}{2}y^p \\
\Rightarrow &\, \left(\frac{1}{2}\right)^p(x+y)^p\leq\frac{1}{2}(x^p+y^p) \\
\Rightarrow &\, (x+y)^p\leq 2^{p-1}(x^p+y^p)
\end{array}$$Thus, for $X\geq0$ and $Y\geq0$, $p\geq1$, we have $$\mathscr{E}\{(X+Y)^p\}\leq2^{p-1}\{\mathscr{E}(X^p)+\mathscr{E}(Y^p)\}\leq2^p\{\mathscr{E}(X^p)+\mathscr{E}(Y^p)\}.$$If $0\leq p\leq1$, consider $f(x,y)=(x+y)^p - (x^p+y^p)$. WLOG, let $x>y\geq0$ and define $t=x/y>1$, we have $$\begin{array}{rl}
f(x,y) & = (x+y)^p - (x^p+y^p) \\
       & = y^p\left(\frac{x}{y}+1\right)^p - y^p\left[\left(\frac{x}{y}\right)^p+1\right] \\
       & = y^p[(t+1)^p - (t^p + 1)] \\
       & = y^p g(t)
\end{array}$$And then $$g'(t) = p[(t+1)^{p-1}-t^{p-1}]<0$$ which implies $g(t)$ is decreasing and $(t+1)^p - (t^p + 1)\leq g(0)=0$. Thus, $$(x+y)^p\leq x^p+y^p.$$ Thus, for $X\geq0$ and $Y\geq0$,
$$\mathscr{E}\{(X+Y)^p\}\leq1\cdot\{\mathscr{E}(X^p)+\mathscr{E}(Y^p)\}\leq2^p\{\mathscr{E}(X^p)+\mathscr{E}(Y^p)\}$$ since $2^p>1$.

$\Box$