ayan.nmath wrote: Let $ABC$ be a triangle with $\angle A=\angle C=30^{\circ}.$ Points $D,E,F$ are chosen on the sides $AB,BC,CA$ respectively so that $\angle BFD=\angle BFE=60^{\circ}.$ Let $p$ and $p_1$ be the perimeters of the triangles $ABC$ and $DEF$, respectively. Prove that $p\le 2p_1.$ Solution. Let $M$ be the midpoint of $AC$. We have $p = (4+2\sqrt{3})\cdot BM$. From the angle conditions, $F$ is the $B-$ Dumpty point of $\triangle BDE$, so $BF^2 = FD\cdot FE$. Further cosine rule in $\triangle DEF$ gives $DE^2 = DF^2 + FE^2 + BF^2\geq 3BF^2$. This gives \[2p_1=2(DE+DF+FE)\geq 2(DE + 2BF)\geq 2BF\left(2+\sqrt{3}\right)\geq 2BM(2+\sqrt{3}) = p\]and we are done.

Let $\omega$ be the circumcircle of $\triangle DEF$, $O$ be the center of $\omega$, $T = \overline{FB} \cap \omega \ne F$, $T'$ be the antipode of $T$ wrt $\omega$. Then $\triangle TDE$ is equilateral. By Ptolemy theorem for quadrilateral $TDFE$, we have $FD + FE = FT$. [asy][asy] size(300); pair T=dir(90),D=dir(-150),E=dir(-30),O=(0,0),F=dir(-120); dot("$T$",T,dir(T)); dot("$D$",D,dir(180)); dot("$E$",E,dir(20)); dot("$O$",O); dot("$F$",F,dir(F)); draw(unitcircle,red); pair B=intersectionpoint(circumcircle(D,O,E),T--F); dot("$B$",B,dir(B)); draw(circumcircle(D,O,E),springgreen); pair Tp=-T; dot("$T'$",Tp,dir(Tp)); draw(F--T--Tp,fuchsia); pair X1=1.1*D-0.1*B,X2=3*D-2*B,X3=D+F-circumcenter(D,T,F),A=intersectionpoint(X1--X2,circumcircle(D,F,X3)); dot("$A$",A,dir(A)); pair C=extension(A,F,B,E); dot("$C$",C,dir(C)); draw(A--B--C--A,brown); draw(F--D--E--F,purple); draw(Tp--F,blue); draw(O--B,blue); draw(D--T--E,blue); [/asy][/asy] Claim: $DE \ge TF \frac{\sqrt{3}}{2} \ge BF \sqrt{3} \ge \frac{AC}{2}$. proof: Since $\angle TFT' = 90^\circ$ (which means $TT' \ge TF$), so $DE = TT' \frac{\sqrt{3}}{2} \ge TF \frac{\sqrt{3}}{2}$. Now, note that $\angle TBO \le 90^\circ$ (as circle with diameter $\overline{TO}$ , $\odot(DOE)$ are tangent to each other at $O$ and $B \in \odot(DOE)$ as $\angle DBE = \angle DOE = 120^\circ$). Now since $\angle TFT' = 90^\circ$ and $O$ is the midpoint of segment $TT'$, so this means that the midpoint of segment $TF$ lies on segment $TB$, which implies $TF \ge 2BF$. Finally, $BF \ge \text{length of } B \text{ altitude in } \triangle ABC = \frac{AC}{2\sqrt{3}}$. This completes the proof of the Claim. $\square$ Hence we obtain that $$ \text{perimeter of } \triangle DEF = DE + (FD + FE) = DE + FT \ge \frac{AC}{2} + \frac{AC}{\sqrt{3}} = \frac{\text{perimeter of } \triangle ABC}{2}$$This completes the proof of the problem. $\blacksquare$