While I was trying to prove some boundedness properties for the Hardy-maximal operator, it was written that
If $q>1, q'=q/(q-1)$, $u$ be a nonincreasing function such that $U(x)=\int_{0}^{x}u(t){\rm d}t,$ and \begin{equation*} \int_{0}^{\infty }\left( \frac{1}{U(x)}\int_{0}^{x}f(t)u(t){\rm d} t\right) ^{q}dx \leq C\int_{0}^{\infty }f^{q}(x)dx, \end{equation*} then by substituting $\ f(t)=\chi _{\left( 0,s\right) }(t)u^{\frac{1}{q-1}}(t)$, where $s$ is a fixed positive number and $\chi _{\left( 0,s\right) }$ is the usual characteristic function, we get that $$\left( \frac{1}{x}\int_{0}^{x}u^{q'}(t)dt\right) ^{1/q'}\leq C\left( \frac{1}{x}% \int_{0}^{x}u(t)dt\right)\tag1$$
Actually, after making the substitution and making some logic simplifications using the monotonicity of $u(\cdot)$, I reached to the following inequality $$\left( \frac{1}{x}\int_{0}^{x}u^{q/(q-1)}(t)dt\right) ^{q-1}\leq C\left(\int_{0}^{x}\frac{x}{\big(xu(t)\big)^q}dt\right)^{-1},$$ and then get stuck to reach inequality (1). Any help related to this is very appreciated.