does anyone know how to solve this problem?? because i've tried a lot and i couldn't solve it...

The pedal triangle of the 1st isodynamic point Q (the intersection of the 3 Apollonius circles inside the triangle $\triangle ABC$) is equilateral. For example: Let D, E, F be the feet of normals from Q to BC, CA, AB. The A-Apollonius circle is a locus of points with the ratio of distances $\frac{AB}{AC}$ from the triangle vertices B, C and similarly for the B- and C-Apollonius circles. Hence $\frac{QB}{QC} = \frac{AB}{AC},$ $\frac{QC}{QA} = \frac{BC}{BA},$ $\frac{QA}{QB} = \frac{CA}{CB}.$ The quadrilateral AEQF is cyclic with the right angles at the vertices E, F, hence its circumradius is $R_A = \frac{AQ}{2},$ so that $EF = 2R_A \sin A = AQ \sin A$ and similarly $FD = BQ \sin B,$ $DE = CQ \sin C.$ Consequently, $\frac{DE}{EF} = \frac{CQ \sin C}{AQ \sin A} = \frac{BC \sin C}{AB \sin A} = 1,\ \ DE = EF$ and similarly EF = FD. From the cyclic quadrilaterals AEQF, BFQD, $\angle AQB = \angle AQF + \angle BQF = \angle AEF + \angle BDF =$ $= 180^\circ - \angle A - \angle AFE + 180^\circ - \angle B - \angle BFD =$ $= 180^\circ - (\angle A + \angle B) + 180^\circ - (\angle AFE + \angle BFD) = \angle C + 60^\circ$ and similarly, $\angle BQC = \angle A + 60^\circ,$ $\angle CQA = \angle B + 60^\circ.$ Let e = DE = EF = FD be the side length of the equilateral pedal triangle $\triangle DEF$. The area S of the triangle $\triangle ABC$ with circumradius R is $S = \frac 1 2 [AQ \cdot BQ\ \sin(C + 60^\circ) + BQ \cdot CQ\ \sin(A + 60^\circ) + CQ \cdot AQ\ \sin(B + 60^\circ)] =$ $= \frac{e^2}{2} [\frac{\sin(C + 60^\circ)}{\sin A \sin B} + \frac{\sin(A + 60^\circ)}{\sin B \sin C} + \frac{\sin(B + 60^\circ)}{\sin C \sin A} ] =$ $= \frac{4R^3e^2}{abc} [\sin A \sin(A + 60^\circ) + \sin B \sin(B + 60^\circ) + \sin C \sin(C + 60^\circ)] =$ $= \frac{R^2e^2}{S} [\frac 1 2 (\sin^2 A + \sin^2 B + \sin^2 C) + \frac{\sqrt 3}{2}(\sin A \cos A + \sin B \cos B + \sin C \cos C)] =$ $= \frac{e^2}{8S} [a^2 + b^2 + c^2 + \frac{4 \sqrt 3 R^2}{2}\ (\sin 2A + \sin 2B + \sin 2C)] =$ $= \frac{e^2}{8S} (a^2 + b^2 + c^2 + 4S \sqrt 3)$ $e = \frac{2S \sqrt 2}{\sqrt{a^2 + b^2 + c^2 + 4S \sqrt 3}}$ Thus the expression on the right side of the inequality in question is precisely the side length of the equilateral pedal triangle $\triangle DEF$ of the 1st isodynamic point Q. Any other equilateral triangle $\triangle D'E'F'$ inscribed in the triangle $\triangle ABC,$ so that $D' \in BC,\ E' \in CA,\ F' \in AB$ is obviously obtained from the equilateral pedal triangle $\triangle DEF$ by a spiral similarity with the center Q and similarity coefficient greater than 1, hence its side e' = D'E' is greater than the side e = DE.

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if $M$ is the fermat point of $ABC$ then the inequality reduces to : $\sum |DE| \cdot |MC|\geq 2S$ this is kind of obvious because the product of the diagonals of a quadrilateral is $\geq$ twice its area...

Not true in general, namely if the 1st Fermat point M is outside the triangle $\triangle ABC,$ say, if $\angle C > 120^\circ.$ Then the inequality is reduced to $|EF| \cdot |MA|+|FD| \cdot |MB|-|DE| \cdot |MC|\geq 2S$ and the "obvious" argument fails. The 1st isodynamic point Q is then also outside the triangle $\triangle ABC$ (inside its circumcircle) and the angle $\angle AQB,$ the convex one, is equal to $\angle AQB = \angle AQF+\angle BQF = 180^\circ-\angle AEF+180^\circ-\angle BDF =$ $\angle A+\angle AFE+\angle B+\angle BFD = 180^\circ-\angle C+120^\circ = 360^\circ-(\angle C+60^\circ)$ so that the concave $\angle AQB = \angle C+60^\circ,$ as before. The area S of the triangle $\triangle ABC$ is still $S = \frac{1}{2}\left[AQ \cdot BQ \sin(C+60^\circ)+BQ \cdot CQ \sin(A+60^\circ)+CQ \cdot AQ \sin(B+60^\circ)\right] =$ $... =\frac{e^{2}}{8S}\left(a^{2}+b^{2}+c^{2}+4S \sqrt 3\right)$ etc.

It is known that among all the equilateral triangles circumscribed in $ \triangle ABC,$ the triangle $ \triangle XYZ$ homothetic to the outer Napoleon triangle of $\triangle ABC$ has the maximum area (side length). Hence, its side is twice the measure of the side $L$ of the outer Napoleon triangle. $ L = \sqrt {\frac {_1}{^6}(a^2 + b^2 + c^2) + \frac {2\sqrt {3}}{3}[\triangle ABC]}.$ For every circumcribed equilateral triangle $ \Delta,$ we can associate an equilateral triangle $ \Delta'$ homothetic to $\Delta$ and inscribed in $ \triangle ABC.$ Thus, by Gergonne-Ann theorem, it follows that $ [\triangle ABC]^2 = [\Delta][\Delta'].$ The area (side L') of $ \Delta'$ will be minimum if the area (side) of $ \Delta$ is maximum. In other words, $ [\triangle ABC]^2 = \frac {_3}{^{16}}(L')^2(4L^2).$ Substituting the value of the side length $ L$ of the Napoleon triangle yields: $ [\triangle ABC]^2 = \frac {_1}{^8}(L')^2(a^2 + b^2 + c^2 + 4\sqrt {3}[\triangle ABC])$ $\Longrightarrow L' = \frac {2\sqrt {2}[\triangle ABC]}{\sqrt {a^2 + b^2 + c^2 + 4\sqrt {3}[\triangle ABC]}}.$

Here is my solution: Denote by $ T$ the Toricelly point of triangle $ ABC$.Let $ DE=EF=FD=x$ Appying Ptolemy's theorem $ S=[AEFT]+[CDTE]+[BDTF]$≤$ \frac{1}{2}(TA.EF+TC.DE+TB.FD)=\frac{x(TA+TB+TC)}{2}$ Hence $ x$≥$ \frac{2S}{TA+TB+TC}=\frac{2S \sqrt 2}{\sqrt{a^2+b^2+c^2+4S \sqrt3}}$

SUPERMAN2 wrote: $ S = [AEFT] + [CDTE] + [BDTF] \ldots$ This only works when the Fermat point lies inside ABC. See post #5

Its well known that orthic triangle has least perimeter of all inscribed triangles. Thus $x>\frac{1}{3}\Sigma a cos\alpha$. So if we show $\frac{1}{3}\Sigma a cos\alpha \ge \frac{2S \sqrt 2}{\sqrt{a^2 + b^2 + c^2 + 4S \sqrt 3}}$ Is this inequality true?

My rough solution is exactly same as done in IMO compendium. Assuming that all the angles of the triangle $ABC$ are less than $120^\circ$. Let the point $T$ be the Torricelli point of the triangle. So, $\angle(AT,EF)=\angle(BT,FD)=\angle(CT,DE)=\theta$. So, $2S= 2(S_{AETF}+S_{BFTD}+S_{CDTE})$ =$(AT+BT+CT)DE\sin\theta\le(AT+BT+CT)DE$ By cosine rule, $AT^2+AT.BT+BT^2=c^2$, $BT^2+CT.BT+CT^2=a^2$, $CT^2+CT.AT+AT^2=b^2$. And $3(AT.BT+BT.CT+CT.AT) =4\sqrt{3}(S_{ATB}+S_{BTC}+S_{CTA}) =4\sqrt{3}S$ Adding all, we get $2(AT+BT+CT)^2=a^2+b^2+c^2+4\sqrt{3}S$.. This combined with the first inequality, gives the result. Now let $\angle{C}\ge 120^{\circ}$. Take T to be the point lying on the same side of $AB$ as $C$ such that $\angle BTC=\angle CTA= 60^{\circ}$ Now we have $2S\ge (AT + BT -CT)DE$ and $2(AT+BT-CT)^2=a^2+b^2+c^2+4\sqrt{3}S$.