Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Sorry if it's a bit murky. I remember posting this one too, and giving a solution I was proud of (can't remember if I actually posted it). However, I can't find that solution.. [Moderator edit: This was at http://www.mathlinks.ro/Forum/viewtopic.php?t=220 .] Let $D',E'$ be the images of $D,E$ through the homothety of center $F$ and ratio $4$. We have to show that $D'E'\le AB+AC$, so it would be enough to show $AE'+AD'\le AB+AC$. Again, we notice that it's enough to show $AD'\le AC\ (*)$. Let $X$ be the vertex of the equilateral triangle $CAX$, lying on the opposite side of $CA$ as $B$. Clearly, $AX=AC$, so $(*)$ is equivalent to $FD'\le FX=FA+FC$ (the last equality is well-known, and it follows from Ptolemy's equality applied to the cyclic quadrilateral $AFCX$) or, in other words, $4FD\le FA+FC$. In terms of areas, this means $4S(FAC)\le S(AFCX)\iff 3S(FAC)\le S(XAC)$, and this is clear since for fixed $XAC$, the area $FAC$ reaches its maximum when $FA=FC$, and in this case we have equality in the above inequality. I think this pretty much ends the proof: we have shown that $4FD\le FX$, which is, as we have shown, equivalent to the initial problem.

I saw this problem these days and I was pretty sure it was an ISL problem. Lets take the equilateral triangles $ ACP$ and $ ABQ$ on the exterior of the triangle $ ABC$. We have that $ \angle{APC} + \angle{AFC} = 180$, therefore the points $ A,P,F,C$ are concyclic. But $ \angle{AFP} = \angle{ACP} = 60 = \angle{AFD}$, so $ D \in (FP)$. Analoguosly we have that $ E \in (FQ)$. Now observe that $ \frac {FP}{FD} = 1 + \frac {DP}{FD} = 1 + \frac {[APC]}{[AFC]}\geq 4$, and the equality occurs when $ F$ is the midpoint of $ \widehat{AC}$. Therefore $ FD \leq \frac {1}{4}FP$, and $ FE \leq \frac {1}{4}FQ$. So, by taking it metrical, we have that: $ DE = \sqrt {FD^{2} + FE^{2} + FD \cdot FE}\leq \frac {1}{4}\cdot \sqrt {FP^{2} + FQ^{2} + FP \cdot FQ} = \frac {1}{4}PQ$ But $ PQ \leq AP + AQ = AB + AC$, and thus the problem is solved.

This post was also a spam and as I am unable to delete this post,i am writing the proof of $\frac{[APC]}{[AFC]} \ge 3$. Note that $(AF-CF)^2 \ge 0 \Rightarrow AF^2+CF^2+AF*CF \ge 3AF*CF \Rightarrow AC^2 \ge 3AF*CF \Rightarrow AP*CP\sin60^{\circ} \ge 3AF*CF\sin120^{\circ} \Rightarrow \frac{[APC]}{[AFC]} \ge 3$.

This is a really nice problem! Thanks to sayantanchakraborty for giving some crucial hints leading to the following solution. Since $\angle BFC, \angle CFA, \angle AFB$ are all equal and sum up to $360^{\circ},$ they must each be equal to $120^{\circ}.$ Construct a point $B'$ outside $\triangle ABC$ such that $\triangle ABB'$ is equilateral. Define point $C'$ analogously. Since $\angle AB'B + \angle AFB = 60^{\circ} + 120^{\circ} = 180^{\circ},$ points $A,B',B,P$ are concyclic. Furthermore, since $\angle B'FB = \angle B'AB = 60^{\circ} = 180^{\circ} - \angle BFC,$ points $C,F,D,B'$ are collinear. I claim that $FB' \geq 4FD.$ This is equivalent to \begin{align*} [\triangle AB'C] & \geq 3[\triangle APC] \\ \iff AB' \cdot B'C \cdot \sin (60^{\circ}) & \geq 3 \cdot AF \cdot CF \cdot \sin (120^{\circ}) \\ \iff AB' \cdot B'C & \geq 3 \cdot AF \cdot CF .\end{align*} By cosine rule on $\triangle AB'C,$ \begin{align*} AC^2 & = AB'^2 + B'C^2 - 2 \cdot AB' \cdot B'C \cdot \cos (60^{\circ}) \\ & = AB'^2 + B'C^2 - AB' \cdot B'C \\ & \geq AB' \cdot B'C , \end{align*} where we have used the trivial inequality $AB'^2 + B'C^2 \geq 2 \cdot AB' \cdot B'C.$ Hence, it suffices to show that \begin{align*} AC^2 & \geq 3 \cdot AF \cdot CF \\ \iff AF^2 + CF^2 - 2 \cdot AF \cdot CF \cos (120^{\circ}) & \geq 3 \cdot AF \cdot CF \\ \iff AF^2 + CF^2 & \geq 2 \cdot AF \cdot CF,\end{align*} which is true. Similar arguments show that $FC' \geq 4FE.$ The rest is obvious. Both the dilations centered at $F$ which map to $B$ to $B'$ and $C$ to $C'$ have ratio at least 4, so $B'C' \geq 4DE.$ By triangle inequality, we have that \[AB'+A'C \geq B'C' \implies AB+AC \geq 4DE. \quad \blacksquare\] For equality to hold, we need $AF=BF=CF,$ that is, the Fermat point must be the circumcenter of $\triangle ABC.$ This is possible iff $\triangle ABC$ is equilateral, because $ \angle AFB = 2 \angle ACB \implies \angle ACB = 60^{\circ}$ and similarly $\angle ABC= 60^{\circ}.$

Erect equilateral triangles $AMC$ and $ANC$ outwardly on the sides of $\triangle ABC$. It is well known that $F \in BM$ and $D \in CN$. $\blacksquare\boxed{\text{Lemma 1}}$ $\frac{1}{FD}=\frac{1}{FA}+\frac{1}{FC}$ and $\frac{1}{FE}=\frac{1}{FA}+\frac{1}{FB}$ . Proof Taking $\angle FAC= \alpha$ and $\angle FCA= \beta$ and using $\alpha + \beta =60$ ,$$\frac{FD}{FA}+\frac{FD}{FC}=\frac{sin(\alpha)}{sin(60+\beta)}+\frac{sin(\beta)}{sin(60+\beta)}=1$$and other part is analogously proved. $\blacksquare\boxed{\text{Lemma 2}}$ $FA+FC \geq 4FD$ and $FA+FB \geq 4FE$ Proof By lemma 1 , $\frac{FA+FC}{FD}= \left(\frac{1}{FA} + \frac{1}{FC} \right) \cdot ( FA+FC ) \geq 4 \implies FA+FC \geq 4FD$ The other part follows analogously. Using lemma 2 $$ AB+AC = AN+AM \geq MN = \sqrt{FN^2+FM^2+FN \cdot FM } = \sqrt{ (FA+FB)^2+(FA+FC)^2+(FA+FB)\cdot(FA+FC)} \geq 4 \cdot \sqrt{FC^2+FD^2+FC \cdot FD}=4DE$$

Construct equilateral triangles $ACX$ and $ABY$ outside of $ABC$, so it's well known that $BFDX$ and $CFEY$ are lines. $YAFB$ is cyclic, so consider the tangent at the point $T$, the antipode of $Y$, labeled as line $\ell$. Note that $d(Y,AB):d(Y,\ell)=3:4$, so then $\dfrac{FE}{FY}\le \dfrac 14$ with equality only when $F=T$, and similarly $\dfrac{FD}{FX}\le \dfrac 14$. Let $M$ and $N$ be the points on segments $FY$ and $FX$, respectively, such that $\dfrac{FM}{FY}=\dfrac 14$ and $\dfrac{FN}{FX}=\dfrac 14$. Then since $FE\le FM$ and $FD\le FN$, $DE=\sqrt{FD^2+FE^2+FD\cdot FE}\le \sqrt{FN^2+FM^2+FN\cdot FM}=MN=\dfrac{XY}{4}\le \dfrac{AB+AC}{4}$, as desired.

cute

Trigonometry is the best weapon. orl wrote: Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \] Clearly $\angle AFB=\angle BFC=\angle CFA=120^o.$ Now, erect equilateral triangles $ABC', BCA', CAB'$ on the sides, externally. Then $AFBC'. AFCB'$ are cyclic. Hence, $\angle C'FA+\angle AFC=\angle C'BA+\angle AFC=60^o+120^o=180^o,$ and so $C, F, C'$ are collinear. We get two more symmetric results and so $F$ is teh Fermat point given by $AA' \cap BB' \cap CC'.$ Claim: $FE: FC' \le 1:4.$ Proof: Ptolemy yields $FC'=FA+FB.$ Hence, it suffices to show $$FA+FB \ge 4FE$$Let $\angle FAB=x.$ Then it suffices to show $$\frac{FA}{FB}+1 \ge \frac{4FE}{FB} \Leftrightarrow \frac{sin(60^o-x)}{sinx}+1 \ge \frac{4sin(60^o-x)}{sin(60^o+x)}$$$$\Leftrightarrow \left(cosx-\sqrt{3}sinx \right)^2 \ge 0$$which is true. $\square$ Similarly we get $FD:FB' \le 1:4$ and so we get $4DE \le B'C' \le AC'+AB'=AB+AC,$ as desired. $\blacksquare$

orl wrote: Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \] The problem is a masterpiece.

Construct points $X, Y, Z$ forming equilateral triangles $\triangle BCX$, $\triangle CAY$, $\triangle ABZ$ sticking out from the triangle. Evidently, $F$ is the intersection of the three circumcircles of these equilateral triangles. Observe additionally that $F \in \overline{AX}$, and in particular $F$ is the concurrence point of $\overline{AX}$, $\overline{BY}$, and $\overline{CZ}$. Note now that $\angle DFE = \angle BFC = 120^\circ$. Thus, we can calculate \begin{align*} DE^2 = DF^2 + FE^2 - 2 \cdot DF \cdot FE \cdot \cos{120^\circ} = DF^2 + FE^2 + DF \cdot FE. \end{align*}As $F$ varies along $(ABZ)$ with length $AB$ fixed, note that the maximum length of $FE$ occurs when $\overline{FZ} \perp \overline{AB}$, and this is also the case for the minimum length of $FZ$. Thus $\frac{FE}{FZ} \leq \frac{1}{4}$. As a result, we know \begin{align*} DE^2 &= DF^2 + FE^2 + DF \cdot FE \\ &\leq \frac{1}{16} (FZ^2 + FY^2 + FZ \cdot FY) \\ &= \frac{1}{16} YZ^2 \\ &\leq \frac{1}{16} (AZ + AY)^2 \\ &= \frac{1}{16} (AB + AC)^2, \end{align*}as desired.

Solved with Alex Zhao, Elliott Liu, Connie Jiang, Groovy (\help), Jeffrey Chen, Nicole Shen, and Raymond Feng. Externally construct equilateral triangles \(ACY\) and \(ABZ\), so that \(B\), \(F\), \(D\), \(Y\) are collinear and \(C\), \(F\), \(E\), \(Z\) are collinear. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,Y,Z,F,D,EE; A=dir(110); B=dir(220); C=dir(320); Y=A+(C-A)*dir(60); Z=B+(A-B)*dir(60); F=extension(B,Y,C,Z); D=extension(B,Y,A,C); EE=extension(C,Z,A,B); draw(D--EE); draw(B--Y,gray); draw(C--Z,gray); draw(circumcircle(A,F,C),linewidth(.3)); draw(circumcircle(A,F,B),linewidth(.3)); draw(C--Y--A--Z--B); draw(A--B--C--cycle,linewidth(.7)); dot("\(A\)",A,dir(105)); dot("\(B\)",B,S); dot("\(C\)",C,S); dot("\(F\)",F,dir(265)); dot("\(Y\)",Y,dir(30)); dot("\(Z\)",Z,dir(150)); dot("\(D\)",D,dir(-5)); dot("\(E\)",EE,dir(210)); [/asy][/asy] Observe that \(FY/FD\ge4\) and \(FZ/FE\ge4\). It follows that \begin{align*} AB+AC=AY+AZ\ge YZ&=\sqrt{FY^2+FZ^2+FY\cdot FZ}\\ &\ge4\sqrt{FD^2+FE^2+FD\cdot FE}=4DE, \end{align*}as needed.

orl wrote: Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \] See also here https://artofproblemsolving.com/community/c6t243f6h2624066_a_refinement_of_imo_shl_2002

First ISL solution in a while. This is the type of problem where if you don't know this property of the Fermat point it's hard to solve (I certainly couldn't do it) and if you know it it's very quick (after I got a hint with the property I solved it within fifteen minutes). It is well-known that $F$ is the Toricelli/1st Fermat point of $\triangle ABC$. It is a well-known property of $F$ that if $ABG$ and $ACH$ are equilateral triangles erected outward from $AB$ and $AC$, respectively, $C,F,G$ are collinear and $AGBF$ is cyclic (similarly $B,F,H$ are collinear and $AHCF$ is cyclic). Notice that as $F$ is on minor arc $AB$, the minimum possible value of $\frac{GE}{EF}$ occurs when $F$ is on the midpoint of the arc, as this maximizes $EF$ and minimizes $EG$. In that case, it is easy to show that $\frac{EG}{EF}=3$ and thus $\frac{FG}{FE}=4$, hence it must be true that $\frac{FG}{FE}\ge 4$ and similarly $\frac{FH}{FD}\ge 4$. Let $FG=\alpha FE$ and $FH=\beta FD$, where $\alpha, \beta \ge 4$. Since $\angle EFD = 120^{\circ}$, by LoC on $\triangle FED$ we get $$ED=\sqrt{FE^2+FD^2-2FE\cdot FD \cos{120^{\circ}}}=\sqrt{FE^2+FD^2+FE\cdot FD}.$$Using LoC on $\triangle FGH$, we get \begin{align*} HG &= \sqrt{FG^2+FH^2-2FG\cdot FH\cos{120^{\circ}}} \\ &= \sqrt{FG^2+FH^2+FG\cdot FH} \\ &= \sqrt{(\alpha FE)^2+(\beta FD)^2+(\alpha FE)(\beta FD)} \\ &= \sqrt{\alpha^2 FE^2+\beta^2 FD^2+\alpha\beta FE\cdot FD} \\ & \ge \sqrt{16FE^2+16FD^2+16FE\cdot FD} \\ &= 4\sqrt{FE^2+FD^2+FE\cdot FD} \\ &= 4ED. \end{align*} By the Triangle Inequality, $AG+AH\ge GH \ge 4ED$, finishing the problem.

Let $B', C'$ be such that $ACB', ABC'$ are equilateral. We have that $C,F,C'$ and $B,F,B'$ are collinear. Claim: $EC' \ge 3EF$ Proof: Note that $\frac{FE}{C'E} = \frac{AF}{AC'} \frac{\sin \angle FAB}{\sin \angle C'AB} = \frac{AF}{c} \frac{2\sin \angle FAB}{\sqrt{3}}$. Let $AF = x$, $BF = y$. Note that $R$, the circumradius of $(AFBC')$, is equal to $\frac{c}{\sqrt{3}}$. We have $2R = \frac{y}{\sin \angle FAB} \implies \sin \angle FAB = \frac{y}{2R} = \frac{\sqrt{3}y}{2c}$. So $\frac{FE}{C'E} = \frac{x}{c} \frac{y}{c} = \frac{xy}{c^2}$. Observe that $c^2 = x^2 + y^2 + xy \ge 3xy \implies \frac{xy}{c^2} \le \frac{1}{3}$. So we have $\frac{FE}{C'E} \le \frac{1}{3} \implies C'E \ge 3FE$, as claimed. $\square$. From the claim, we have $DE \le \frac{B'C'}{4} \le \frac{AB' + AC'}{4} = \frac{AB+AC}{4} \implies AB + AC \ge 4DE$, as desired. $\blacksquare$

Note that ∠AFB = ∠BFC = ∠CFA = 120 so making regular triangles with bases AB and AC is a good move. Let S and K be outside ABC such that ABS and ACK are regular triangles. Note that AFBS and AFCK are cyclic. Let O1,O2 be reflections of F across AB and AC. FE/ES = [AFB]/[ABS] = [AO1B]/[ASB] ≤ 1/3 so FS ≥ 4FE. Same way we can prove FK ≥ 4FD. so SK ≥ 4DE and SK ≤ AS + AK = AB + AC. we're Done.

Try the refinement https://artofproblemsolving.com/community/c6t243f6h2624066_a_refinement_of_imo_shl_2002

What. Let $G$ be on $FD$ extended such that $\angle AGC=60^\circ.$ Let $H$ be on $FE$ extended such that $\angle AHB=60^\circ.$ Note that $AFCG$ is cyclic. Also, $\angle AFD=60^\circ$ and $\angle CFD=60^\circ$ so $\angle CAG=\angle ACG=60^\circ.$ Thus, $ACG$ is equilateral. Similarly, $AHB$ is equilateral. Now, $AB+AC\ge HG.$ Since $\angle HFG$ is obtuse, it suffices to show $HF\ge 4EF$ and $GF\ge 4DF$ to prove that $HG\ge 4ED.$ Note that $\triangle HFA\sim \triangle HAE$ by AA so $\frac{HE}{HF}=\left(\frac{HE}{HA}\right)^2\ge \left(\frac{\sqrt{3}}{2}\right)\ge \frac{3}{4},$ which implies the result that $HF\ge 4EF$. Similarly, $GD\ge 4DF.$ Now, WLOG suppose the parallel line through $E$ parallel to $HG$ lies outside of $\triangle EDF.$ Then this line intersects $FG$ at $J.$ $HG\ge EJ\ge ED$ as desired.

Construct equilateral triangles $\triangle ABX$ and $\triangle ACY$ outside of $\triangle ABC$ and note that $AXBF$ and $AYCF$ are cyclic. It's easy to see that $FX\ge 4FE$ and $FY\ge 4FD$ so by the Law of Cosines we easily obtain $XY\ge 4DE$ so that \[AB+AC=AX+AY\ge XY\ge 4DE.\]We are done. $\blacksquare$

Really cute orl wrote: Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \] Draw points $K, L$ such that $AKB$ and $ALC$ are equilateral triangles. Clearly, $AFCL, AFBK$ are cyclic quads, and $\angle AFL=\angle ACL=180^{\circ}-\angle AFB=60^{\circ}$ implies $B, F, D, L$ are collinear. Similarly, $C, F, E,$ and $K,$ are collinear. Now $AB+AC=AK+AL \ge KL$ so it suffices to show that $DE \leq \tfrac{1}{4} KL$. We will show that $FE \leq \tfrac{1}{4}FK$ and $FD \leq \tfrac{1}{4}FL$. It suffices to prove the following two lemmas to finish: Lemma 1. Point $W$ lies on arc $\widehat{YZ}$ of the circumcircle of equilateral triangle $XYZ$ not containing $X$ and line $XW$ meets $YZ$ at point $T$. Then $WX \geq 4WT$. Proof: Indeed, it is enough to show $XT \ge 3WT$. Now $XT$ is larger than the $X$-median of $\triangle XYZ$ and $WT$ is smaller than the length it achieves when $W$ is antipodal to $X$. For rigour, this follows as $XW \cdot XT$ is fixed by shooting lemma. When $W$ is antipodal, equality is achieved, proving the lemma. Lemma 2. In obtuse triangle $XYZ$ with obtuse angle at $X$, points $Y_1, Z_1$ lie on rays $XY, XZ$ such that $XY \geq 4XY_1$ and $XZ \geq 4XZ_1$. Then $YZ \geq 4Y_1Z_1$. Proof: Scale by a factor of $4$ to assume $XY_1 \leq XY$ and $XZ_1 \le XZ$. Now $Y_1Z_1<Y_1Z$ as $\angle Y_1Z_1Z>\angle Y_1XZ>90^{\circ}$ and $Y_1Z<YZ$ as $\angle YY_1Z>\angle YXZ>90^{\circ}$, so $Y_1Z_1<YZ$ unless $Y_1=Y$ and $Z_1=Z$, proving the claim. Finally, by combining Lemma 1 and Lemma 2 in triangle $FKL$ for points $D$ and $E$, the conclusion follows.

r poblem Construct equilateral triangles $ABX$ and $ACY$ so that both are not in the of $ABC$. Then it is well known that $A$, $F$, $X$ and $B$, $F$, $Y$ are collinear, and moreover $AXBF$, $AYCF$ are cyclic. Now we can obtain $FX\ge 4FE$, $FY\ge 4FD$ and hence by LOC $XY \ge 4DE$. To finish, \[AB+AC=AX+AY\ge XY\ge 4DE\]as desired. $\square$

Point $F$ is $\textbf{First Fermat Point}$ and construction can easily be found from the theorem thonk:/

Note $F$ is the Fermat Point. Thus, let $X$ and $Y$ be points such that $\triangle ABX$ and $\triangle ACY$ are equilateral triangles lying outside $\triangle ABC$. Claim. $FX \geq 4 \cdot FE$ and $FY \geq 4 \cdot FD$. Proof. We will prove this with area ratios. Note, $AB^2 = AF^2 + FB^2 + AF \cdot FB$, by Law of Cosines. \begin{align*} \dfrac{[AXB]}{[AFB]} = \dfrac{\tfrac{\sqrt{3}}{4} \cdot AB^2}{\tfrac{1}{2} \cdot AF \cdot FB \cdot \sin\left(120^{\circ}\right)} = \dfrac{AF^2 + FB^2 + AF \cdot FB}{AF \cdot FB} = \dfrac{AF}{FB} + \dfrac{FB}{AF} + 1 \geq 3. \end{align*}Thus, $[AXB] \geq 3 \cdot [AFB]$, thus $XE \geq 3 \cdot EF$ and $FX \geq 4 \cdot FE$. Then, $FY \geq 4 \cdot FD$ follows, as desired. $\blacksquare$ Note that by triangle inequality this clearly implies $XY \geq 4 \cdot DE$. But, $AB + AC = AX + AY \geq XY$, by triangle inequality. Therefore $AB + BC \geq 4 \cdot DE$, as desired.

Clearly $\angle AFB = \angle BFC = \angle CFA = 120^\circ.$ Erect equilateral triangles $\triangle ACP$ and $\triangle ABQ$ outside of $\triangle ABC.$ Let $AF = x$ and $FB = y.$ Observe that $AFBQ$ is cyclic as $\angle AQB + \angle AFB = 60^\circ + 120^\circ = 180^\circ.$ Thus by Ptolemy on $AFBQ,$ we get $FQ = FA + FB = x + y.$ Since $\triangle FAE \sim \triangle FQB$ (by simple angle-chasing), we get $FE \cdot FQ = FA \cdot FB,$ so $FE = \frac{FA \cdot FB}{FA + FB} = \frac{xy}{x+y}.$ Therefore, $$\frac{FE}{FQ} = \frac{xy}{(x+y)^2} \le \frac{xy}{4xy} = \frac{1}{4}$$by AM-GM on the denominator. Similarly, $\frac{FD}{FP} \le \frac{1}{4}.$ Therefore, $$AB + AC = AQ + AP \ge QP \ge 4DE,$$as desired.

By similar triangles and angle bisector theorem, we may compute \[EF = AE \cdot \frac{BF}{AB} = AB \cdot \frac{AF}{AF+BF} \cdot \frac{BF}{AB} = \frac{AF \cdot BF}{AF+BF}.\]Now let $a = AF$, $b = BF$, and $c = CF$, and observe that $EF = \frac{ab}{a+b} \leq \frac{a+b}4$ while $FD \leq \frac{a+c}4$. From here, it is very much feasible to directly expand $(AB+AC)^2 \geq 16 DE^2$ using Law of Cosines, but here is a comparatively nicer finish. Erect equilateral triangles $BCX$, $ACY$, and $ABZ$ outside triangle $ABC$ such that $F = \overline{AX} \cap \overline{BY} \cap \overline{CZ}$, and note that $EF \leq \frac 14 FZ$, et cetera. So \[DE^2 = EF^2+DF^2 + DE \cdot EF \leq \frac{FZ^2+FY^2 + FZ \cdot FY}{16} = \frac{ZY^2}{16} \leq \frac{(AB+AC)^2}{16}\]as needed. Remark: For some reason, this felt quite hard for G2.