let me post my solution: let $ P , Q$ the intersection of the paralle with $ AB$ and the sides $ AC$ and $ AB$ respectively , let $ x = AP,y = BQ, PQ = z$,the quadrilateral $ APQB$ has an incircle so $ c + z = x + y$ and by thales $ \frac {b - x}{b} = \frac {a - y}{y} = \frac {z}{c}$ , solving this system we find easily : $ PQ = z = \frac {c(a + b - c)}{a + b + c} ,CP = b - x = \frac {b(a + b - c)}{a + b + c}$, and $ CQ = \frac {a(a + b - c)}{a + b + c}$. let $ m = \frac {(a + b - c)}{a + b + c}$. the inradius of triangle $ CPQ$ is $ \sqrt {\frac {(ma + mb - mc)(ma + mc - mb)(mb + mc - ma)}{4(ma + mb + mc)}} = mr$ ,where $ r$ is the inradius of triangle $ ABC$. and with the same way for the other triangles . the sum of areas is $ \pi(m^2 + n^2 + p^2 + 1)r^2$ where $ p = \frac {(a - b + c)}{a + b + c}$ and $ q = \frac {( - a + b + c)}{a + b + c}$. which equal to: ${ \pi([\frac {[a + b - c}{a + b + c}]^2 + [\frac {a - b + c}{a + b + c}]^2 + [\frac { - a + b + c}{a + b + c}]^2 + 1)\frac {(a + b - c)(a + c - b)(b + c - a)}{4(a + b + c}}$ $ = \pi[\frac {(a^2 + b^2 + c^2)(a + b - c)(a + c - b)(b + c - a)}{(a + b + c)^3}$.

Well, actually, I think this is a routine problem. I'm using similiarities, and the ratio between the base and the height of 1 small triangle and 1 big triangle.

dyd wrote: Given the triangle $ ABC$ with sidelengths $ a,b,c$ . Tangents to incircle of $ ABC$ which are parallel with triangle's sides form three small triangles ( each small triangle has one vertex of $ ABC$ ). Prove that the sum of area of incircles of these three small triangles and the area of incircle of triangle $ ABC$ is equal to $ \frac {\pi (a^{2} + b^{2} + c^{2})(b + c - a)(c + a - b)(a + b - c)}{(a + b + c)^{3}}$ Proof. Denote the incircle $ w=C(r)$ of $ \triangle ABC$ , the points $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ where the circle $ w$ touches the triangle $ ABC$ , the points $ M\in (AB)$ , $ N\in (AC)$ for which $ MN\parallel BC$ and $ MN$ is tangent to $ w$ , the incircle $ w_1=C(r_1)$ and the semiperimeter $ p_1$ of $ \triangle AMN$ . Observe that $ \triangle AMN\sim\triangle ABC$ and $ p_1=p-a$ and $ \frac {r_1}{r}=\frac {p_1}{p}$ $ \implies$ $ r_1=\frac {r(p-a)}{p}$ . In conclusion, $ r^2+\sum r_1^2=r^2+\left(\frac rp\right)^2\cdot\sum (p-a)^2=$ $ \frac {r^2}{p^2}\cdot\left[p^2+\sum (p-a)^2\right]=$ $ \frac {r^2}{p^2}\cdot\sum a^2=$ $ \frac {S^2}{p^4}\cdot\sum a^2=$ $ \frac {(p-a)(p-b)(p-c)}{p^3}\cdot\sum a^2=$ $ \frac {(b+c-a)(c+a-b)(a+b-c)}{(a+b+c)^3}\cdot\left(a^2+b^2+c^2\right)$ .