From the Ptolemy Inequality on $\Box BEFC$, we have $EF \cdot BC + BF \cdot CE \ge CF \cdot BE \ge AF \cdot EC + AE \cdot BF + BC \cdot FE \implies BF \cdot CE \ge AF \cdot CE + AE \cdot BF$ Let $AF:FB = x:(1-x), AE:EC=y:(1-y)$. We have that $(1-x)(1-y) \ge x(1-y) + y(1-x) \implies 3xy-2x-2y+1 \ge 0$ Also, since $\frac{AG}{GD} - 1 = \frac{AF}{FB} + \frac{AE}{EC} - 1= \frac{x}{1-x} + \frac{y}{1-y} - 1 = \frac{2x+2y-3xy-1}{(1-x)(1-y)} \le 0$, we have that $AG \le GD$ $\blacksquare$

By Ptolemy's inequality on quadrilateral $BFEC$ and the hypothesis give $$BC\cdot FE+BF\cdot EC\geq FC\cdot BE\geq AF \cdot EC+ BF \cdot AE+BC \cdot FE$$and implies \begin{align} BF\cdot EC\geq AF\cdot EC+BF\cdot FE \end{align} Van Aubel's Theorem: Given a triangle with concurring cevians $AD, BE,$ and $CF$ at $G$, $\frac{AG}{GD}=\frac{AF}{FB}+\frac{AE}{EC}$. Proof: $\frac{AG}{GD}=\frac{AD}{GD}-1=\frac{[ABC]}{[BGC]}-1=\frac{[ABGC]}{[BGC]}$. Also, $\frac{AF}{FB}+\frac{AE}{EC}=\frac{[AGC]}{[BGC]}+\frac{[AGB]}{[BGC]}=\frac{[ABGC]}{[BGC]}$ so Van Aubel's Theorem is proven.$\Box$ By Van Aubel's Theorem and (1), $$\frac{AG}{GD}=\frac{AF}{FB}+\frac{AE}{EC}=\frac{AF\cdot EC+FB\cdot AE}{BF\cdot EC}\leq 1$$so $AG\leq GD$, as desired. $\blacksquare$