Spivak's book states first this definition of Convex Function:
Definition 1:
A function $f$ is convex on an interval, if for all $a$ and $b$ in the interval, the line segment joining $(a, f(a))$ and $(b, f(b))$ lies above the graph of $f$.
Now, this means that the stright line defined by the function $g(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$ is such that $f(x)<g(x)$. But this is equivalent to say that $\frac{f(x)-f(a)}{x-a}<\frac{f(b)-f(a)}{b-a}$. This is the reason why then it states an equivalent definition.
Definition 2:
A function $f$ is convex on an interval if for $a$, $x$, and $b$ in the interval with $a<x<b$ we have $\frac{f(x)-f(a)}{x-a}<\frac{f(b)-f(a)}{b-a}$.
Now, analogously, I can take the same stright line defined by $g$ but now described by the function $G(x)=\frac{f(b)-f(a)}{b-a}(x-b)+f(b)$ in such a way that, if $f$ is convex by definition $(1)$, then $f(x)<G(x)$ and therefore
RESULT: $\frac{f(x)-f(b)}{x-b}>\frac{f(b)-f(a)}{b-a}$ because $x-b <0$.
My question is then, how can I get this result from definition $2)$?.
Best Answer
Start with definition 2, $\frac{f(x)-f(a)}{x-a}<\frac{f(b)-f(a)}{b-a}$. Cross multiply (both denominators positive) and break-up: $$\left[f(x)-f(a)\right](b-a)<\left[f(b)-f(a)\right](x-a)$$ $$\Rightarrow f(x)(b-a)-f(a)b+f(a)a < \left[f(b)-f(a)\right]x-f(b)a+f(a)a $$ Cancel out $f(a)a$ from both sides, move $f(b)a$ from RHS to LHS and subtract from both sides $b\left[f(b)-f(a)\right]$: $$\Rightarrow f(x)(b-a)-f(a)b + f(b)a-b\left[f(b)-f(a)\right] < \left[f(b)-f(a)\right]x-b\left[f(b)-f(a)\right] $$
Simplify and collect terms :$$f(x)(b-a) - f(b)(b-a) < \left[f(b)-f(a)\right](x-b)$$
$$\Rightarrow \frac{f(x)-f(b)}{x-b}>\frac{f(b)-f(a)}{b-a}$$ because $x-b$ is negative.