A convex function has a lower bound? Suppose that $f=f(x)$ is strictly convex for $x\in\mathbb{R}$, i.e. there exists $\epsilon>0$ such that $f''(x)\geq\epsilon>0$ for $x\in\mathbb{R}$. Does there exist $\delta>0$ such that $f(x)\geq \delta$ for $x\in\mathbb{R}$?

A function satisfying the condition given in this question is called "strongly convex". 
There is a more general definition of strong convexity which applies to functions that may not be differentiable:  A function $f:\mathbb R^N \to \bar{\mathbb R}$ is strongly convex (with parameter $\epsilon > 0$) if and only if the function $h(x) = f(x) - \frac{\epsilon}{2} \|x\|_2^2$ is convex.  
It would be nice to give a version of this result that doesn't require $f$ to be differentiable.
Assume that $f:\mathbb R^N \to \mathbb R \cup \{\infty\}$ is strongly convex (with parameter $\epsilon$) but not necessarily differentiable.  The function $h(x) = f(x) - \frac{\epsilon}{2} \|x\|_2^2$ is convex and never equal to $-\infty$, therefore $h$ has an affine minorant: there exists $m \in \mathbb R^N$ and $b \in \mathbb R$ such that $h(x) \geq \langle m, x \rangle + b$ for all $x \in \mathbb R^N$.  It follows that 
\begin{equation}
f(x) \geq \frac{\epsilon}{2} \|x\|_2^2 + \langle m, x \rangle + b
\end{equation}
for all $x \in \mathbb R^N$.  The quadratic function on the right has a lower bound, so $f$ has a lower bound.
