I am actually reading an article where the authors used the following whithout mentioning any proof of it $$\int_{0}^{\infty} f^\lambda(t)d(t^\lambda) \le \lambda\int_{0}^{\infty} \left(\int_0^t f(\tau)d\tau\right)^{\lambda -1}f(t) dt = \left(\int_{0}^{\infty} f(t)dt\right)^\lambda $$
Where $\lambda\ge 1$, the expression $d(t^\lambda)$ stand for $\lambda t^{\lambda-1}dt$ and the function $f$ is a nonnegative nonincreasing and integrable function on $(0,\infty).$
My first Guess is that this inequality must be related to some convex inequalities (which I do not know): Because for $\lambda\ge 1$ the function $x\mapsto x^\lambda$ is convex and this function is very linked to this problem.
Now the equality $$\lambda\int_{0}^{\infty} \left(\int_0^t f(\tau)d\tau\right)^{\lambda -1}f(t) dt = \left(\int_{0}^{\infty} f(t)dt\right)^\lambda $$ is rather easy to check since by setting $$F(t) = \int_0^t f(\tau)d\tau$$
we end up with
$$\lambda\int_{0}^{\infty} \left(\int_0^t f(\tau)d\tau\right)^{\lambda -1}f(t) dt = \lambda\int_{0}^{\infty} F(t)^{\lambda -1}F'(t) dt \\= F^\lambda(\infty)-F^\lambda(0)= \left(\int_{0}^{\infty} f(t)dt\right)^\lambda $$
Any reference, idea, help or proposal is very welcome.
Since $f$ is non-increasing, we have
$$\int_0^t f(\tau) \, d\tau \geq \int_0^t f(t) \, d\tau = t f(t).$$
Thus,
$$\begin{align*} \int_0^{\infty} f^{\lambda}(t) \, d(t^{\lambda}) &= \lambda \int_0^{\infty} (t f(t))^{\lambda-1} f(t) \, dt \\ &\leq \lambda \int_0^{\infty} \left( \int_0^t f(\tau) \, d\tau \right)^{\lambda-1} f(t) \, dt. \end{align*}$$
The following example shows that the monotonicity of $f$ is a crucial assumption.
Example: Consider
$$f(t) := 1_{[1,2]}(t),$$
then
$$\int_0^{\infty} f^{\lambda}(t) \, d(t^{\lambda}) = \bigg[t^{\lambda} \bigg]_{t=1}^{2} = 2^{\lambda}-1$$
and
$$\left( \int_0^{\infty} f(t) \, dt \right)^{\lambda} = 1^{\lambda} = 1;$$
therefore we have
$$\int_0^{\infty} f^{\lambda}(t) \, d(t^{\lambda}) =2^{\lambda}-1>1 = \left( \int_0^{\infty} f(t) \, dt \right)^{\lambda} \qquad \text{for all $\lambda>1$}.$$