let $D\in BC,E\in AC,\F\in AC$ you can prove easily,that minimum area of $DEF$ will occur when $DEF$ be a equalateral triangle. now let $BD=x$ hence $DE^{2}=\frac{7}{4}x^{2}-2x+1$ and minimum of $DE$ is when $x=\frac{4}{7}\rightarrow DE=\sqrt{\frac{3}{7}}$

How do you know that $CE=\frac{\sqrt{3}x}{2}$ and $BF=1-\frac{x}{2}$?

Iso92 wrote: How do you know that $CE=\frac{\sqrt{3}x}{2}$ and $BF=1-\frac{x}{2}$? let $H$ be feet of perpendicular from $D$ to $AB$ rotate $AB$ around $D$ by angle $60^{\circ}$ this line intersect $AC$ at $E$,let $H'$ be rotation of point $H$ . it's clear that $DH'\perp BC\Longrightarrow DH'=CA=\frac{\sqrt{3}}{2}x\ \ , \ H'E=HF=1-x\Longrightarrow \ BF=1-\frac{x}{2}$ .

"you can prove easily that minimum area of DEF will occur when be a equalateral triangle. " That's not right. If F -> B, E -> C and no matter where D is on BC, the area of DEF -> 0.

Vo Duc Dien wrote: "you can prove easily that minimum area of DEF will occur when be a equalateral triangle. " That's not right. If F -> B, E -> C and no matter where D is on BC, the area of DEF -> 0. I think the claim of Amir.S is correct.Suppose $DEF$ is a triangle with $DF>DE>EF$ and $DF$ is the minimum of the longest side of all possible inscribable triangles.Then by assumption there exists a point $E'$ on $(AC)$ such that $DF=DE'$.Consider the triangle $DE'F$.Now slide $DE'F$ by moving $E'$ and $F$ along $AC$ and $AB$ such that $DF,DE'$ become shorter at the same rate and $E'F$ becomes longer gradually.At a point of time $DE'F$ will become equilateral with sides shorter than $DF$,a contradiction.The same reasoning goes for isosceles triangles.