The first inequality is
$\int_{0}^{\infty} x^af(x)dx \leq a(\int_{0}^{\infty} xf(x)dx) $ for $0<a<1$ and
$\int_{0}^{\infty} x^af(x)dx \geq a(\int_{0}^{\infty} xf(x)dx)$ for $a>1$
The second inequality is
$\int_{0}^{\infty} x^af(x)dx \leq (\int_{0}^{\infty} xf(x)dx)^a $ for $0<a<1$ and
$\int_{0}^{\infty} x^af(x)dx \geq (\int_{0}^{\infty} xf(x)dx)^a$ for $a>1$
where $f(x)$ is a probability density function and $x>0$.
Could the first or second inequlaity be able to prove?
Any help would be appreciated.
The first inequalities are not true: suppose $f(x)=1$ for $x\in [0,1]$ and $0$ otherwise. Then for $0<a<1$, $$ \int_0^{\infty} x^af(x)dx=\int_0^{1} x^adx\geq \int_0^1 xdx=\int_0^{\infty} xf(x)dx> a\int_0^{\infty} xf(x)dx $$ and similarly, for $a>1$, $$ \int_0^{\infty} x^a f(x)=\int_0^1 x^a dx\leq \int_0^1 xdx =\int_0^{\infty} x f(x)dx<a\int_0^{\infty} xf(x)dx. $$
The second inequalities however are true, and follow immediately from Jensen's Theorem: when $0<a<1$ the map $x\mapsto x^a$ is concave and so thinking of $f(x)$ as the PDF of some nonnegative random variable $X$: $$ \int_0^{\infty} x^af(x)dx=\mathbb{E}[X^a]\leq \mathbb{E}[X]^a=\bigg(\int_0^{\infty} xf(x)dx\bigg)^{\alpha}. $$ Similarly, when $1<a$, the map $x\mapsto x^a$ is convex and so $$ \int_0^{\infty} x^af(x)dx=\mathbb{E}[X^a]\geq \mathbb{E}[X]^a=\bigg(\int_0^{\infty} xf(x)dx\bigg)^{\alpha}. $$