Let $C',F'$ be the reflections of $C,F$ respectively in $BE$. We have $GB+GA=GC',\ HD+HE=HF'$, so $GB+GA+HD+HE+GH=C'G+GH+HF'\ge C'F'=CF$, Q.E.D.

Take points $ L$ and $ K$ inside $ ABCDEF$ such that in triangles $ ABL$ and $ DEK$ we have: $ AB=BL=LA$ and $ DE=DK=EK$. Obviously, $ ALBG$ and $ DHEK$ are cyclic. By ptoleme theorem result, $ AG+GB=GL$ and $ HD+HE=HK$. We get convex hexagon $ DBLAEK$, which is equal to convex hexagon $ DCBAFE$, as $ AB=BC=CD=DB=BL=LA$; $ AF=FE=ED=DK=KE=AE$; $ \angle DCB=\angle EFA=\angle DKE=\angle BLA=1/3\pi$; So, $ CF=KL$ and $ AG+GB+GH+HD+HE=GL+GH+KH$. Left to prove, that $ GL+GH+KH\geq KL$, which is right as $ KL$ is the least way from point $ K$ to point $ L$, Q.E.D.

Can this problem can be generalized to any points $G$ and $H$ in the plane of $ABCDEF$ by Ptolemy's inequality?

SnowEverywhere wrote: Can this problem can be generalized to any points $G$ and $H$ in the plane of $ABCDEF$ by Ptolemy's inequality? The answer is Yes. Let $P$ be a point on the plane of equilateral $\triangle ABC$. $PA, PB, PC$ form a triangle (maybe a degenerate triangle). This theorem is known as Pompeiu's Theorem. Let's go back to the IMO problem. We are constructing outer equilateral triangles $\triangle G'AB$ and $\triangle H'DE$. It has proven $G'H' = CF$ by many times. We will prove $GA+GB+GH+DH+HE \geq GG' + GH + HH' \geq G'H' = CF$. We only need $GA+GB \geq GG'$ and $DH+HE \geq HH'$. That's what Pompeiu's Theorem states.

We construct two points $P$ and $Q$ outside $ABCDEF$ such that triangles $ABP$ and $DEQ$ are equilateral triangles. Then by Ptolemy's theorem, we have $PG = AG + BG$ and $QH = HE + HD$. Thus, $AG + BG + GH + HD + HE = PG + GH + HQ \geq PQ$ (1) Now consider $CBAFED$ and $PBDQEA$. They have all the corresponding angles equal and their lengths also equal. Thus they are congruent, and we have $PQ = CF $----- (2) Thus the inequality follows from (1) and (2) EDIT: Points P and Q outside ABCDEF, and such that $ABP \cap ABCDEF = AB $ and $DEQ \cap ABCDEF = DE$

Since \(\triangle BCD\) and \(\triangle EFA\) are equilateral, we have \(AB=BD\) and \(AE=ED\), so quadrialteral \(ABDE\) is a kite. Now consider the reflections \(C'\) and \(F'\) of \(C\) and \(F\) over \(\overline{BE}\). Since \(G\in(C'AB)\) and \(H\in(F'DE)\), \[CF\le C'G+GH+HF'=(GA+GB)+GH+(HD+HE).\]This completes the proof. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A,D,B,EE,C,F,Cp,Fp,G,H; A=(-1,0); D=(1,0); B=(0,1.2); EE=(0,-2); C=B+(D-B)*dir(60); F=EE+(A-EE)*dir(60); Cp=reflect(B,EE)*C; Fp=reflect(B,EE)*F; G=circumcenter(Cp,A,B)+circumradius(Cp,A,B)*dir(-30); H=circumcenter(Fp,D,EE)+circumradius(Fp,D,EE)*dir(135); draw(Cp--G--H--Fp,gray+linewidth(.4)+dashed); draw(circumcircle(Cp,A,B),gray+linewidth(.3)); draw(circumcircle(Fp,D,EE),gray+linewidth(.3)); draw(A--B--D--EE--cycle); draw(B--C--D); draw(EE--F--A); draw(A--Cp--B,linewidth(.4)+dashed); draw(D--Fp--EE,linewidth(.4)+dashed); draw(A--G--B,linewidth(.3)); draw(D--H--EE,linewidth(.3)); dot("\(A\)",A,W); dot("\(B\)",B,N); dot("\(C\)",C,NE); dot("\(D\)",D,E); dot("\(E\)",EE,S); dot("\(F\)",F,SW); dot("\(C'\)",Cp,NW); dot("\(F'\)",Fp,SE); dot("\(G\)",G,E); dot("\(H\)",H,W); [/asy][/asy]

Solution. Let $C'$, $F'$ be the reflections of $C$, $F$ over $BC$, respectively. It's easy to see that $\triangle AEF$, $\triangle BCD$, $\triangle ABC'$, $\triangle DEF'$ are all equilateral. Also by the angle conditions $DHEF'$, and $BGAC'$ are cyclic quadrilaterals. Using Ptolemy's Theorem on those quadrilaterals $\implies GA+GB=GC'$, and $HE+HD=HF'$. Now we have that, \[AG + GB + GH + DH + HE= GC'+GH+HF' \geq C'F'=CF.\blacksquare \]

Solved with OronSH, qwerty123456asdfgzxcvb, Zhaom, and polishedhardwoodtable at 1 AM. Let $Z$ be the point such that $ABZ$ is equilateral and $Z$ lies opposite $D$ and $E$ wrt $AB$ or something. Repeat the construction for $EDY$, this time on the opposite side. By symmetry, we have that $(F,Y)$ and $(C,Z)$ are reciprocal pairs under the plane involution of reflection across $BE$. Therefore $YZ$ = $CF$. Now we know by say, Van Schooten's theorem that $AG + GB = GZ$ and $EH + HD = HY$. Particularly we have that the inequality is equivalent to $HY + GH + GZ \ge YZ$, which is obvious by two applications of the triangle inequality.