Hexagon and incircle of triangle inequality. > Consider the three lines tangent to the incircle of a triangle $ABC$ which are parallel to the sides of the triangle; these, together with the sides of the of the triangle, for a hexagon $T$. Prove that $$\text{the perimeter of $T$} \leq \dfrac{2}{3} \text{the perimeter of $\triangle{ABC}.$}$$ There is an interesting way to think of this question. We can think of the incircle as an excircle to three different triangles. I am not sure if that helps or if there is a name for such a geometrical shape, but I think it might help in solving the question. Other than that there seem to be a lot of similar triangles to use. ![enter image description here](

Referring to the illustration in the question, we have $DE = Dz + Ey$. Using the two analogous equalities for $FG$ and $HI$, we find that the perimeter of $T$ is $2DE + 2FG + 2HI$.

In triangle $ABC$, let $a, b, c$ be the sides, let $p$ be the semi-perimeter, let $S$ be the area, let $r$ be the radius of the incircle, and let $h_a$ be the altitude from $A$.

Triangle $ADE$ is similar to $ABC$ in the ratio $$1 - \frac{2r}{h_a} = 1 - \frac{2S/p}{2S/a} = 1 - \frac{a}{p}.$$ Therefore $2DE = 2a(1 - a/p) = \frac{a}{p} (b + c - a)$. Then the perimeter $P$ of the hexagon is $$P = \frac{a}{p} (b + c - a) + \frac{b}{p} (a + c - b) + \frac{c}{p} (a + b - c),$$ and the problem is to show that $$(a + b + c)^2 - 2(a^2 + b^2 + c^2)=pP \leq \frac{4}{3}p^2 = \frac{1}{3}(a + b + c)^2.$$ This follows easily from the Cauchy-Schwarz inequality applied to the vectors $(a,b,c)$ and $(1,1,1)$.