Proving that the graph of a convex function lies above its asymptote Suppose that a function $f: \left( a, \infty \right) \to \mathbb{R}$, where $a \in \mathbb{R}$, is twice differentiable and convex and that $f$ has an asymptote at $+ \infty $. I want to prove that the graph of $f$ never goes below the asymptote. The problem seems to be trivial but I can't solve it. Does anyone have any ideas?
 A: Suppose the asymptote is $y = c x + d$.  This says that the function $g(x) = f(x) - c x - d \to 0$ as $x \to +\infty$.  Note that $g(x)$ is also convex.
Suppose $g(t) < 0$ for some $t$.  Since $g(x) \to 0$ as $x \to +\infty$, there is some $s > t$ such that $g(s) > g(t)/2$.  By convexity, for $x > s$ we have
$$ g(x) \ge g(s) + (x-s) \frac{g(s)-g(t)}{s-t} > \frac{g(t)}{2} - \frac{(x-s)g(t)}{2(s-t)} $$
and the right side goes to $+\infty$ as $x \to +\infty$, contradicting the assumption that $g(x) \to 0$ as $x \to +\infty$.
A: Let $y=mx+b$ be the asymptote.
Suppose that $f(x_0)<mx_0+b$ for some $x_0\in (a,\infty)$. By the property of an asymptote, for $\epsilon:=|f(x_0)-(mx_0+b)|$, there exists $L$ such that for all $x>L$, we have $|f(x)-(mx+b)|<\epsilon$. Thus for $x>\max\{L,x_0\}$, we have that $$mx+b-f(x)\le |f(x)-(mx+b)|<|f(x_0)-(mx_0+b)|=mx_0+b-f(x_0)$$
and so $$\frac{f(x)-f(x_0)}{x-x_0}>m $$
It follows that $f'(\xi)>m$ for some $\xi\in(x_0,x)$. As $f'$ is non-decreasing, it follows that $f(x)\ge f(\xi)+(x-\xi)f'(\xi)$ for all $x>\xi$. As $f'(\xi)>m$, we can find $x_1$ such that $f(x)>mx+b+1$ follows for all $x>x_1$, contradicting the asymptote property.
