A friend and I were trying to figure out this problem from our assignment.

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> Prove that on an open $I$, a geometrically convex function $f(x)$ is continuous. 
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To better assist the audience, it is best I give you the definition of a geometrically convex function from the book, "Introduction to Analysis," by Arthur Mattuck.

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> Let $f(x)$ be defined on any type of interval I. For any subinterval $[a,b] \subset I$, we let $P:(a,f(a))$ and $Q:(b,f(b))$ be the two points of the graph lying over the endpoints of the interval. We say $f(x)$ is geometrically convex on $I$ if the graph on f(x) lies on or below the chord PQ, for all $[a,b]\subset I$ An equivalent analytic formulation of this is $\frac{f(x)-f(a)}{x-a}\leq\frac{f(b)-f(x)}{b-x}$, for all $a<x<b$ in $I.$ 
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How would we go about pursuing this problem? I would figure a way to approach it would be to show $\lim\_{\bigtriangleup x\rightarrow0^-}\frac{\bigtriangleup y}{\bigtriangleup x}$ exists at each point of $I$, and some how deduce $\lim\_{\bigtriangleup x\rightarrow0^-}\bigtriangleup y=0$. Is this correct? If so, how would I proceed?