Question: I'm solving some exercises about integrals but I can't solve this one, can someone explain the step-to-step please?

Let $\phi:[a,b]\rightarrow\mathbb{R}$ be a function with a second derivative, where $f(0)=0$ and $f(\pi)=3$, calculate $\int\_{0}^{\pi}[f(t)+f''(t)]sin(t)dt$

Answer: 
Let $p>1$, $d\mu(t)=\frac{dt}{t}$ and
$$
\Phi(t)=\frac{1}{t}\int\_0^t\frac{f(s)}{s}\,ds.
$$
We note that we can write $\Phi(t)=\int\_0^1\frac{f(ty)}{ty}\,dy$, and thus, by Fubini's theorem,
$$
\begin{aligned}
\int\_0^{+\infty}\Phi(t)^p\,d\mu(t)&=\int\_0^{+\infty}\biggl[\int\_0^1\frac{f(ty)}{ty}\,dy\biggr]\cdot\Phi(t)^{p-1}\,d\mu(t)\\
&=\int\_0^1\biggl[\int\_0^{+\infty}\frac{f(ty)}{ty}\cdot\Phi(t)^{p-1}\,d\mu(t)\biggr]\,dy.
\end{aligned}
$$
Next, we invoke the Hölder inequality on the inner integral, and find that the integral in the right-hand side is bounded by (here $1/p+1/q=1$, so $q(p-1)=p$)
$$
\int\_0^1\biggl[\int\_0^{+\infty}\Bigl(\frac{f(ty)}{ty}\Bigr)^p\,d\mu(t)\biggl]^{1/p}\cdot\biggl[\int\_0^{+\infty}\Phi(t)^p\,d\mu(t)\biggr]^{1/q}\,dy
$$
By letting $s=ty$ in the integral in the left square bracket, we find that it is independent of $y$. But this means that the integrand is completely independent of $y$, and thus the $y$ integral just evaluates to $1$, and what we got is really
$$
\biggl[\int\_0^{+\infty}\Bigl(\frac{f(s)}{s}\Bigr)^p\,d\mu(s)\biggl]^{1/p}\cdot\biggl[\int\_0^{+\infty}\Phi(t)^p\,d\mu(t)\biggr]^{1/q}
$$
If we now divide by
$$
\biggl[\int\_0^{+\infty}\Phi(t)^p\,d\mu(t)\biggr]^{1/q}
$$
we find that
$$
\biggl[\int\_0^{+\infty}\Phi(t)^p\,d\mu(t)\biggr]^{1-1/q}\leq
\biggl[\int\_0^{+\infty}\Bigl(\frac{f(s)}{s}\Bigr)^p\,d\mu(s)\biggl]^{1/p}.
$$
Raising both sides to $p$, we get the inequality we was about to prove.

This proof was in fact @JanG's idea (given in a private discussion we had). It is also worth mentioning that one can also prove the inequality by using the Minkowski inequality and some corollaries to it. The details are given in the book *Real Analysis* by Folland, see Theorem 6.20 and Exercise 29 on the page after.
I can prove this inequality if $f (x) /x$ were non-decreasing.

All we need to prove is $$\int\_{0}^{t} \frac {f (s)} {s} ds \leqslant f(t).$$ Take $g (x) = f (x) / x$ wherever defined, then we should prove $$\int\_{0}^{t} g (s) ds \leqslant t g(t).$$ Since $g$ is non-decreasing, we are done.