Tangents of a Strictly Convex Fuction Let $f:\mathbb R \rightarrow \mathbb R$ be differentiable and strictly convex.  Is it true that $x,y \in \mathbb R$ and $x \neq y$ imply
\begin{equation}
f'(x)(y-x) <f(y) - f(x)
\end{equation}
If so, how can we prove it?
 A: Pick an arbitrary $x,y \in \mathbb R$ such that $x \neq y$.
Since $f$ is convex (strict convexity not required yet), we can conclude that
\begin{equation}
f'(x)(z-x) \leq f(z) - f(x), \ \ \forall z \in \mathbb R.
\end{equation}
In particular, the inequality holds for $z = tx + (1-t)y$, where $t \in (0,1)$:
\begin{equation}
f'(x)(tx + (1-t)y-x) \leq f(tx + (1-t)y) - f(x).
\end{equation}
We will now analyze both sides of this inequality.
For the LHS, we can simply rearrange terms to obtain
\begin{equation}
f'(x)(tx + (1-t)y-x) = (1-t)f'(x)(y-x).
\end{equation}
For the RHS, we can apply the strict convexity of $f$ and rearrange terms to obtain
\begin{equation}
f(tx + (1-t)y) - f(x)< t(f(x) + (1-t)f(y) - f(x) = (1-t)(f(y) - f(x)).
\end{equation}
Bringing it all together, we have
\begin{equation}
(1-t)f'(x)(y-x) < (1-t)(f(y) - f(x)).
\end{equation}
Since $t \in (0,1)$, this is equivalent to the desired result:
\begin{equation}
f'(x)(y-x) < f(y) - f(x).
\end{equation}
A: Yes this is true. An equivalent condition for convexity is that $f(x)$ lies above its tangent at every point. This is precisely the statement that $f'(x) < \frac{f(y) - f(x)}{y-x}$, i.e. the derivative should be smaller than the slope of the secant line.
