Proof that Right hand and Left hand derivatives always exist for convex functions. Definition 
A function $f$ is convex on an interval if for $a,x, \text{and} \;b$ in the interval with $a\lt x\lt b$, we have
$$\frac{f(x)-f(a)}{x-a}\lt \frac{f(b)-f(a)}{b-a}.$$
While reading the proof that if $f$ is convex on some open interval containing $a$, then $f_+'(a)$ and $f_{-}'(a)$ always exist. The proof first shows that $[f(a+h)-f(a)]/h$ is decreasing as $h\to 0^+$, so 
$$f_+'(a)=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}=\operatorname{inf}\{\frac{f(a+h)-f(a)}{h}:h\gt 0\}.$$
The proof states that this inf exists because each quotient $\frac{f(a+h)-f(a)}{h}$ for $h\gt 0$ is greater than any one such quotient for $h'\lt 0$. 
However, I'm struggling to show this inequality, that for any $h\gt 0$ and $h'\lt 0$, we have
$$\frac{f(a+h')-f(a)}{h'}\lt \frac{f(a+h)-f(a)}{h}.$$
I would greatly appreciate it if anyone can help me prove this inequality.
 A: Note: I don't assume strict convexity, so I use weak inequalities. For strictly convex $f$, all the inequalities between expressions of the form $\frac{f(y) - f(x)}{y-x}$ are strict.
We have
\begin{align}
\frac{f(b) - f(a)}{b-a} &= \frac{\bigl(f(b) - f(x)\bigr) + \bigl(f(x) - f(a)\bigr)}{b-a}\\
&= \frac{b-x}{b-a}\cdot\frac{f(b) - f(x)}{b-x} + \frac{x-a}{b-a}\cdot\frac{f(x) - f(a)}{x-a}\\
&= \lambda\cdot\frac{f(b) - f(x)}{b-x} + (1-\lambda)\cdot\frac{f(x) - f(a)}{x-a},
\end{align}
where $0 < \lambda = \frac{b-x}{b-a} < 1$.
By the convexity condition
$$\frac{f(x) - f(a)}{x-a} \leqslant \frac{f(b) - f(a)}{b-a}$$
it then follows that
$$\frac{f(b) - f(a)}{b-a} \leqslant \frac{f(b) - f(x)}{b-x}.$$
That holds for any three points $a < x < b$ in the interval on which $f$ is convex.
Now apply the chain of inequalities to $u = a+h',\, v = a,\, w = a+h$ to obtain the desired
$$\frac{f(a+h') - f(a)}{h'} \leqslant \frac{f(a+h) - f(a)}{h}.$$
A: I'm going to suggest you think geometrically here (which includes drawing a picture): Suppose $A,B,C$ are points in the plane with $A$ to the left of $B$ and $B$ to the left of $C.$ The first inequality you cite is of the form $s(A,B) \le s(A,C),$ where $s$ denotes slope. How could this be true unless $B$ lies on or below the line through $A$ and $C?$ And if $B$ satisfies that, doesn't it follow that $s(A,B) \le s(B,C)?$ The latter inquality is what you are trying to prove. So this is transparent geometrically; you "just" have to translate this picture into a bit of analysis.
