Show convex function is increasing in both variables of difference quotient using alternative definition of convexity Let $\phi$ be a function that satisfies 
$$\frac{\phi (t) - \phi (s)}{t - s} \leq \frac{\phi (u) - \phi (t)}{u - t}$$
where $s < t < u$. 
Is it possible to directly use this definition of convexity to prove that $\phi$'s difference quotients are increasing in each variable, i.e., 
$$\frac{\phi (u) - \phi (s)}{u - s} \leq \frac{\phi (u) - \phi (t)}{u - t}$$
and 
$$\frac{\phi (t) - \phi (s)}{t - s} \leq \frac{\phi (u) - \phi (s)}{u - s}$$
where again $s < t < u$.
Background: $\phi$ is convex and using the usual definition of convexity the proof is fairly direct. So I'm curious if there's a direct proof from this definition of convexity.
 A: All three relations are equivalent, and equivalent to
$$
\begin{vmatrix} 
1 & 1 & 1 \\
s & t & u \\
\phi(s) & \phi(t) & \phi(u)
\end{vmatrix} \ge  0
$$
for $s < t < u$. In fact this determinant is equal to each of 
$$\begin{vmatrix} 
1 & 0 & 0 \\
* & t-s & u-t \\
* & \phi(t) - \phi(s) & \phi(u)-\phi(t)
\end{vmatrix} 
= ( \phi(u)-\phi(t)) (t-s) - (\phi(t) - \phi(s))(u-t)
$$
$$
\begin{vmatrix} 
0 & 0 & 1 \\
s -u & t - u & * \\
\phi(s) - \phi(u) & \phi(t) - \phi(u) & *
\end{vmatrix}
= (\phi(u) - \phi(t))(u-s) - (\phi(u) - \phi(s))(u-t)
$$
$$
\begin{vmatrix} 
1 & 0 & 0 \\
* & t-s & u-s \\
* & \phi(t)-\phi(s) & \phi(u)-\phi(s)
\end{vmatrix} 
= (\phi(u)-\phi(s))(t-s)) - (\phi(t)-\phi(s))(u-s)
$$
as can be seen by elementary column operations on the determinant.
A: Let $S$ be the point $(s, \phi (s))$, and define $T$ and $U$ respectively. The condition given is that the gradient of $ST$ is less than that of $TU$, where the $x$-coordinate of $T$ is between the other two. Draw the triangle $STU$, and interpret all difference quotients as gradients. The result should now be clear.
