All triples are $a>0$ and $b^2=4ac$, or $a=b=c=0$. Let $g(t)=at^2+bt+c$. Then: 1) $g(t)\geq0$ holds for all $t\in\mathbb{R}$. Otherwise if $g(t_0)<0$ for some $t_0$, we get $|f(x+t_0)-f(x)|\leq g(t_0)<0$, which is impossible) 2) $\min g(t)=0$. Otherwise if $\min g(t)=m>0$, take $f(x)=\begin{cases}{m,}&{x=0,}\\{0,}&{x\neq0.}\end{cases}$ We get $|f(x)-f(y)|\leq m\leq g(x-y)$. From 1)2) we know that $a>0$ and $b^2=4ac$, or $a=b=c=0$. It remains to prove that $f$ has to be constant under this condition. The $a=0$ case is trivial. Now let $a>0$ and we know $$|f(x)-f(y)|\leq a(x-y-r)^2$$for some $r\in\mathbb{R}$. If $r\neq 0$, let $x=y+r$ and then we know that $f$ is periodic with period $r$. Hence $$|f(x)-f(y)|=|f(x+r)-f(y)|\leq a(x+r-y-r)^2=a(x-y)^2.$$So we only need to consider the case $r=0$. This time we have $$\left|\dfrac{f(x)-f(y)}{x-y}\right|\leq a|x-y|.$$By taking the limit $y\rightarrow x$, we know that $f'(x)=0$ for all $x\in\mathbb{R}$, which means that $f$ is constant.

I think we can not use derivative here. But we can end solution in next way We want to prove that $|f(x)-f(y)| \leq a(x-y)^2$ with $a>0$ has only const solution Let $x-y=nd$ where $n$ is natural $|f(x)-f(y)| =|f(y+nd)-f(y)| \leq |f(y+nd)-f(y+(n-1)d)|+|f(y+(n-1)d)-f(y+(n-2)d)|+...+|f(y+d)-f(y)| \leq and^2=\frac{a(x-y)^2}{n}$ So by taking limit $n\to +\infty$ we get $f(x)=f(y)$