I couldn't think that there is no combinatorics geometry in IGO Intermediate this year.

Sketch. Let $DE$ intersect $BC$ at $X$, $CE$ intersect $DA$ at $Y$, then note that $XBFD$ and $YAFC$ are cyclic, so use PoP and the parralel lines $AD$ and $BC$ with Thales theorem to finish (at the end the ineq is equivalent with $AE/BE+BE/AE=>2$ which is obvious AM-GM)

I gave a BAD solution to this( some trigonometric inequalities involved) but short

Let $\angle FBC=\angle EDC=\angle AED=\alpha$ and $\angle FAD=\angle ECD=\angle CEB=\beta$. We have $\angle ADC=\angle ABC, \angle EDC=\angle FBC$. Hence, $\angle ADE=\angle ABF$. Let $\angle ADE=\angle ABF=\theta$. Similarly, let $\angle BAF=\angle BCE=\delta$. Then, $\angle AFD=\angle FAB=\delta$ and $\angle BFC=\angle ABF=\theta$. $0<\alpha, \beta, \theta, \delta$ and $\alpha+\beta+\theta+\delta=180^\circ$. Hence; $\sin \alpha, \sin \beta, \sin \theta, \sin \delta>0$. Suppose that $|AB|<2|BC|$. By the Law of Sines in the triangles $AED$ and $BEC$, we can find that $\frac{|AE|}{\sin \theta}=\frac{|AD|}{\sin \alpha}$ and $\frac{|EB|}{\sin \delta}=\frac{|BC|}{\sin \beta}$. Thus, $2|BC|>|AB|=|AE|+|EB|=\frac{|AD|\cdot \sin\theta}{\sin \alpha}+\frac{|BC|\cdot \sin\delta}{\sin \beta}=|BC|\left(\frac{\sin\theta}{\sin \alpha}+\frac{\sin\delta}{\sin \beta}\right)\Rightarrow 2>\frac{\sin\theta}{\sin \alpha}+\frac{\sin\delta}{\sin \beta}.$ Similarly, by the Law of Sines in the triangles $ADF$ and $BCF$, we can find that $2>\frac{\sin\alpha}{\sin \theta}+\frac{\sin\beta}{\sin \delta}.$ Hence, $4>\frac{\sin\theta}{\sin \alpha}+\frac{\sin\delta}{\sin \beta}+\frac{\sin\alpha}{\sin \theta}+\frac{\sin\beta}{\sin \delta}=\left(\frac{\sin\theta}{\sin \alpha}+\frac{\sin\alpha}{\sin \theta}\right)+\left(\frac{\sin\delta}{\sin \beta}+\frac{\sin\beta}{\sin \delta}\right)\ge AM-GM\ge 2+2=4.$ Contradiction. Thus, $|AB|\ge 2|BC|$.

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I gave a solution using similarity followed by AM GM during the contest too(very similar/same as the official solution). All in all, I think this was a tough one according to me, as I hadn't seen AM GM in geo anywhere before this.

My sol:- Reflection of $DE$ across angle bisector of $\angle ADC$ , reflection of $CE$ across angle bisector of $\angle DCB$ meet at a point, say, $F'$ on $AB$ by symmetry. By PoP $AF'*AE=AD^2=BC^2=BE*BF'=(AB-AE)(AB-AF')$ This imply $AB=AF'+AE$.So, $(AF'-AE)^2\geq0$ imply $AB^2=(AF'+AE)^2\geq4*AF'*AE=4BC^2$. So, done.

Yay, my first problem in a contest. Hope it was well received

Nice problem. Let $\angle FBC=\angle FDE=\alpha$, $\angle ECD=\angle FAD=\beta$, $\angle ECB=\phi$, $\angle EDA=\theta$ Since $ABCD$ a parallelogram. $\angle BCD=\angle BAD \text{ but } \angle ECD=\angle FAD \Rightarrow \angle BCE = \angle EAF=\phi$ $\angle CBE=\angle CDA \text{ but } \angle FBC=\angle EDC \Rightarrow \angle FBE = \angle EDA=\theta$ Since $AB\parallel CD\Rightarrow \angle BEC=\angle ECF=\beta \text{ and } \angle EAF = \angle AFD=\phi$ By Law of Sines in $\triangle BEC \text{ and } \triangle EAD$ \[\frac{AB}{BC}=\frac{BE}{BC}+ \frac{EA}{AD}= \frac{\sin\theta}{\sin \alpha}+ \frac{\sin\phi}{\sin \beta}\] By Law of Sines in $\triangle BFC \text{ and } \triangle FAD$ \[\frac{CD}{BC}=\frac{AB}{BC}=\frac{FC}{BC}+ \frac{FD}{AD}= \frac{\sin\alpha}{\sin \theta}+ \frac{\sin\beta}{\sin \phi}\] Finishing: \[\left(\frac{AB}{BC}\right)^2=\left(\frac{\sin\theta}{\sin \alpha}+\frac{\sin\phi}{\sin \beta}\right)\left(\frac{\sin\alpha}{\sin \theta}+ \frac{\sin\beta}{\sin \phi}\right)\underbrace{\geq}_{\text{By AM-GM}}4\cdot\sqrt{\frac{\sin\theta}{\sin \alpha}\frac{\sin\phi}{\sin \beta}\frac{\sin\alpha}{\sin \theta}\frac{\sin\beta}{\sin \phi}}=4\] \[\left(\frac{AB}{BC}\right)^2\geq 4\] \[\frac{AB}{BC}\geq 2\Rightarrow AB\geq 2BC,\text{As desired.}\]