Sufficient condition for decreasing function $\phi:(0,\infty)\to (0,\infty)$ to obey $\int_x^\infty\phi(t)\mbox{d}t\le Cx\phi(x)$, with C = constant 
Question. What is a simple characterization of strictly decreasing functions $\phi:[1,\infty) \to (0,\infty)$, with $\int_1^\infty \phi(t)\mbox{d}t \lt \infty$ and $\lim_{x \to \infty}\phi(x) = 0$, such that
$$
\int_x^{\infty} \phi(t)\mbox{dt} \le C x\phi(x)\,\forall x \ge 1,
$$
for some positive constant $C$ which only depends on $\phi$ ?

For example, the functions $x \mapsto x^{-1-\gamma}$ (with $\gamma \gt 0$) and $x \mapsto e^{-x}$ verify the above condition.
Note. In the absense of a succinct characterization, I'm fine with a sufficient condition which covers a diverse choises of $\phi$ at once.
 A: Denote $F(x) = \int_x^{\infty}\phi(t)\mbox{dt}$, then
$$\begin{align}
&\Longleftrightarrow F(x) \le - Cx F'(x)  \\
&\Longleftrightarrow F'(x)+\frac{1}{Cx}F(x) \le 0   \\
&\Longleftrightarrow \frac{F'(x)}{F(x)}+\frac{1}{Cx} \le 0  \qquad \text{as } F(x)>0 \\
&\Longleftrightarrow \left(\ln\left(F(x)x^{\frac{1}{C}}  \right)\right)' \le 0 \\
&\Longleftrightarrow \ln\left(F(x)x^{\frac{1}{C}}  \right) \quad \text{is non-increasing}\\
&\Longleftrightarrow F(x)x^{\frac{1}{C}}  \quad \text{is non-increasing}  \tag{1}  \\
&\Longleftrightarrow x^{\frac{1}{C}}\int_x^{+\infty}\phi(t)dt  \quad \text{is non-increasing}    \\
&\Longleftrightarrow x^{1+\frac{1}{C}}\int_1^{+\infty}\phi(xt)dt  \quad \text{is non-increasing}  \tag{2}  \\
\end{align}$$
There are two solutions.
First solution: The necessary and sufficient condition is:  there exists a $\gamma >0$ such that $F(x)x^{\gamma}$ is a non-increasing function. We determine the function $\phi(x)$ as follows:

*

*Step 1: Define a function $p(x)$ and a $\gamma >0$ satisfying: $p(x)$ is non-increasing function and $x^{-\gamma}p(x)$ is convexe (the convexity is necessary and sufficient for the strictly decreasing of $\phi(x)$).
For example: $p(x) = x^{\gamma}e^{-x}$ with $\gamma<1$. We have $x^{-\gamma}p(x) = e^{-x}$ is convexe


*Step 2: With the $\gamma$ and $p(x)$ already defined in step 1, assume that $F(x)x^{\gamma} = p(x) \Longleftrightarrow F(x) = x^{-\gamma}p(x)$ then
$$\color{red}{\phi(x) = -F'(x) = -\left(x^{-\gamma}p(x)  \right)'}$$
We have $\phi(x)$ is strictly decreasing (as $\left(x^{-\gamma}p(x)  \right)''>0$).
For example, if $p(x) = x^{\gamma}e^{-x}$ then $\phi(t) = e^{-x}$
Second solution: from $(2)$, we deduce that the sufficient solution is: there exists a $\gamma >0$ such that for all $t>1$ $x \mapsto x^{1+\gamma}\phi(xt)$ is a non-increasing function
Remark: I thought that the second solution would be more interesting, but now I think the first one is much better. Indeed, the first solution is the necessary and sufficient condition and it's easy to create the function $\phi(x)$.
A: This is not a full solution but a general method, reminiscent of Gronwall's inequality, that may shed some light into the OP's question.
Setting $h(x)=\int^\infty_x\phi$, we obtain that
$$-h'(x)=\phi(x)\geq \frac{1}{cx}h$$
or equivalently
$$h'+\frac{1}{cx}h\leq0$$
multiplying by $x^{1/c}$ gives
\begin{align}
x^{1/c}\Big(h'+\frac{1}{xc}h\big)=(x^{1/c}h)'\leq0\tag{1}\label{one}
\end{align}
Integrating over $(t,T)$
$$t^{-1/c}T^{1/c}h(T)\leq h(t)\leq ct\phi(t)$$
That is
\begin{align}
\phi(t)\geq\frac{1}{c}T^{1/c}h(T)t^{-1/c-1}\tag{2}\label{two}
\end{align}
If $\phi$ is such that $T^{1/c}\int^\infty_T\phi\xrightarrow{T\rightarrow\infty}\alpha$ (the limit exists since $x\mapsto x^{1/c}h(x)$ is nonnegative and monotone nonincreasing), then
\begin{align}
\phi(t)\geq \frac{\alpha}{c}t^{-1/c-1}\tag{3}\label{three}
\end{align}
Conversely, suppose $\phi$ is non negative, continuous, integrable in $(1,\infty)$, such that $x\mapsto x^{1/c}\int^\infty_x\phi$ is monotone non increasing, then
$$\big(x^{1/c}\int^\infty_x\phi\big)'=x^{1/c}\Big(-\phi(x)+\frac{1}{cx}\int^\infty_x\phi\Big)\leq0$$
Hence
$\int^\infty_x\phi \leq cx\phi(x)$, and by the first part of the posting, it satisfies \eqref{three}.

Continuity is to avoid issues of differentiability of $x\mapsto \int^\infty_x\phi$ and avoid passing to almost surely type of arguments.
