$(F\ast G)(x)\leq F(x)G(x)$ Consider the CDFs $F$ and $G$ on $\mathbb R_+$.
How can I show that $$(F\ast G)(x)\leq F(x)G(x),\ \ \ x \in \mathbb R_+$$
$|(F\ast G)(x)|\leq\int_{\mathbb R_+}|F(x-y)G(y) |dy$ but how do I continue? Do we have that $F$ is bounded? (because it is a cdf, so it is bounded by $1$?) Then I could write $...\leq \|F\|_\infty\|G\|_1=\|G\|_1$
?
How would a 'probabilistic' (I don't know if this is the correct word) look like? (i.e. using appropriate random variables and probabilities?
Edit:
Apparently, for distribution functions, the convolution is defined as $$\int_{-\infty}^{\infty}F(x-y)dG(y)$$ which is $$\int_{-\infty}^{\infty}F(x-y)G'(y)dy$$ for absolute continuous $G$.
Instead of $$\int_{-\infty}^{\infty}F(x-y)G(y)dy$$
Okay, so this should be the solution:
$\int_{-\infty}^{\infty}F(x-y)dG(y)=\int_{0}^{x}F(x-y)dG(y)\leq\int_{0}^{x}F(x)dG(y)=F(x)G(x),\quad x \geq 0$
 A: Before dealing with the problem itself, we may need one more setting.
Because we have talked about convolution, i.e. $\int_{\mathbb R^+}F(x-y)G(y)dy$, we need that $F$ is also define on $\mathbb R^-$. The reason is that, for any fixed $x\in\mathbb R^+$, when $y$ ranges over $\mathbb R^+$, $x-y$ will range over $(-\infty, x)$. Since your functions $F$ and $G$ are CDFs, we may assume that they are zero on $\mathbb R^-$(i.e. we identify the zero extensions of them with themselves).
With this setting in mind, we can get started now.
The statement in the question is not true, at least for large x.
Counter-Example:
First, because $F$ and $G$ are CDFs, we know that $\lim_{x\to\infty} F(x)=1=\lim_{x\to\infty} G(x)$. And since CDFs are non-decreasing, there exists a real number $a$ s.t. $F(a)>0.9$ and $G(a)>0.9$ .
Pick an $x\in\mathbb R^+$ so large that $x-(a+100)>a.$ So,  $\forall y\in(a,a+100),$ $x-y>a.$
Then, $$\int_a^{a+100} F(x-y)G(y)dy>\int_a^{a+100} F(a)G(a)dy=F(a)G(a)\cdot 100 >81.$$
Because $F$ and $G$ are non-negative, we have that, $$\int_{\mathbb R^+}F(x-y)G(y)dy >\int_a^{a+100} F(x-y)G(y)dy>\int_a^{a+100} F(a)G(a)dy=F(a)G(a)\cdot 100>81.$$
But $F(x)\le 1$ and $G(x)\le 1,$ which implies that for this $x$, we have that $(F\ast G)(x) >81>1 \ge F(x)G(x),$ i.e. $(F\ast G)(x) > F(x)G(x).$

A weaker result (maybe also the best result) we can prove will be: $(F\ast G)(x)\le xG(x)F(x) ,\,\forall x\in \mathbb R^+.$
Fix $x \in \mathbb R^+.$
$|(F\ast G)(x)|\leq\int_{\mathbb R^+}|F(x-y)G(y) |dy=\int_{A\cup B}|F(x-y)G(y) |dy,$ where $A:= \mathbb{R^+} \cap \{x\geq y\}$ and $B:= \mathbb{R^+} \cap \{x\lt y\}$.
On $B$, $F=0,$ and so the integral over $B$ is zero.
On A, because $\{ x\ge y\}$, we see $G(y)\le G(x), \forall y \in A.$ This is the non-decreasing property of CDF.
So, $\int_A|F(x-y)G(y) |dy\le G(x)\int_A|F(x-y) |dy\le G(x)\int_A|F(x)|dy$, where the last inequality holds because CDF is non-negative and non-decreasing and $x-y\in [0,x)$ on $A$.
Now, $G(x)\int_A|F(x)|dy\le G(x)|F(x)|\int_A dy= G(x)|F(x)|x=G(x)F(x)x$.
So we have shown that, $(F\ast G)(x)\le xG(x)F(x).\,\forall x\in \mathbb R^+.$
$$\tag*{$\blacksquare$}$$
Remark 1: By the estimate we proved in the answer, we can see that the statement in the problem is true for $x\in(0,1].$ But as the counter-example shows, it is not true for large $x$.
Remark 2: Actually we don't need any absolute sign in the proof, because every CDF and the convolution of CDFs are non-negative. I didn't realize this when I started writing the proof. I am lazy to change it but you should be aware of this.
