Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x \in (a,b)$ $$f(a)<A$$ $$f(x)+f'(x)<A$$ Show that $f(x)<A$ for all $x \in (a,b)$.
My attempt: Generate a function $h(x)$ such that $$h(x)=e^xf(x)$$ $$\Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$ So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d \in (a,b)$ such that $$h(c)-h(d)=h'(\frac{c+d}{2})(c-d)$$ $$e^cf(c)<Ae^{\frac{c+d}{2}}(c-d)+Ae^d$$ $$f(c)<Ae^{\frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{\frac{1}{2}}(c-d)+1]$$ It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.
I don't see what you are bringing $c$ and $d$ into it. Let $c\in(a,b)$. Then $$h(c)-h(a)=\int_a^c h'(t)\,dt <A\int_{a}^c e^t\,dt=Ae^c-Ae^a.$$ Then $$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$ Then $f(c)<A$.