Let $f(x)$ and $g(x)$ be probability density functions defined on $[0,\infty)$.
We have that $$ \int_{0}^\infty x^n g(x) dx \le \int_0^{\infty} x^n f(x) dx \quad {for} \quad n=0,1,\ldots \infty.$$
Can one prove that $$ \int_{0}^\infty x^{n+1} g(x^2) dx \le \int_0^{\infty} x^{n+1} f(x^2) dx \quad {for} \quad n=0,1,\ldots \infty? \tag{1}$$
Substituting $x=y^2$ in the top equation, I get $$ \int_{0}^\infty y^{2n} g(y^2) dy^2 \le \int_0^{\infty} y^{2n} f(y^2) dy^2 \quad {for} \quad n=0,1,\ldots \infty.$$ which simplifies to $$ \int_{0}^\infty y^{2n+1} g(y^2) dy \le \int_0^{\infty} y^{2n+1} f(y^2) dy \quad {for} \quad n=0,1,\ldots \infty,$$ which proves equation (1) for $n$ even.
Can one prove equation (1) for the case where $n$ is an odd number, or is there an altogether simpler way?