Given triangle $ ABC$ with sidelengths $ a,b,c$. Tangents to incircle of $ ABC$ that parallel with triangle's sides form three small triangle (each small triangle has 1 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}\plus{}b^{2}\plus{}c^{2})(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)}{(a\plus{}b\plus{}c)^{3}}$

(hmm,, looks familiar, isn't it? :wink: )
1. Denote the incircle of $\triangle ABC$ as $w = C(r)$, where $r$ is the inradius. Let the points where the incircle touches the sides $BC$, $CA$, and $AB$ be $D$, $E$, and $F$ respectively.
2. Consider the tangents to the incircle that are parallel to the sides of $\triangle ABC$. These tangents form three smaller triangles, each sharing one vertex with $\triangle ABC$. For instance, let $M \in AB$ and $N \in AC$ such that $MN \parallel BC$ and $MN$ is tangent to $w$.
3. Denote the incircle of $\triangle AMN$ as $w_1 = C(r_1)$ and its semiperimeter as $p_1$. Since $\triangle AMN \sim \triangle ABC$, we have $p_1 = p - a$, where $p$ is the semiperimeter of $\triangle ABC$.
4. The similarity ratio between $\triangle AMN$ and $\triangle ABC$ gives $\frac{r_1}{r} = \frac{p_1}{p}$. Therefore, $r_1 = \frac{r(p - a)}{p}$.
5. The area of the incircle of $\triangle ABC$ is $\pi r^2$. The area of the incircle of $\triangle AMN$ is $\pi r_1^2 = \pi \left(\frac{r(p - a)}{p}\right)^2 = \pi \frac{r^2 (p - a)^2}{p^2}$.
6. Similarly, for the other two smaller triangles, we have $r_2 = \frac{r(p - b)}{p}$ and $r_3 = \frac{r(p - c)}{p}$, and their areas are $\pi r_2^2 = \pi \frac{r^2 (p - b)^2}{p^2}$ and $\pi r_3^2 = \pi \frac{r^2 (p - c)^2}{p^2}$.
7. Summing the areas of the incircles of the three smaller triangles and the incircle of $\triangle ABC$, we get:
   \[
   \pi r^2 + \pi r_1^2 + \pi r_2^2 + \pi r_3^2 = \pi r^2 + \pi \frac{r^2 (p - a)^2}{p^2} + \pi \frac{r^2 (p - b)^2}{p^2} + \pi \frac{r^2 (p - c)^2}{p^2}
   \]
   \[
   = \pi r^2 \left(1 + \frac{(p - a)^2 + (p - b)^2 + (p - c)^2}{p^2}\right)
   \]
8. Using the identity $(p - a)^2 + (p - b)^2 + (p - c)^2 = a^2 + b^2 + c^2 - p^2$, we have:
   \[
   \pi r^2 \left(1 + \frac{a^2 + b^2 + c^2 - p^2}{p^2}\right) = \pi r^2 \left(\frac{a^2 + b^2 + c^2}{p^2}\right)
   \]
9. Since $r = \frac{A}{p}$, where $A$ is the area of $\triangle ABC$, we have $r^2 = \frac{A^2}{p^2}$. Substituting this in, we get:
   \[
   \pi \frac{A^2}{p^2} \left(\frac{a^2 + b^2 + c^2}{p^2}\right) = \pi \frac{A^2 (a^2 + b^2 + c^2)}{p^4}
   \]
10. Using Heron's formula, $A = \sqrt{p(p - a)(p - b)(p - c)}$, we have $A^2 = p(p - a)(p - b)(p - c)$. Substituting this in, we get:
    \[
    \pi \frac{p(p - a)(p - b)(p - c) (a^2 + b^2 + c^2)}{p^4} = \pi \frac{(p - a)(p - b)(p - c) (a^2 + b^2 + c^2)}{p^3}
    \]
11. Finally, substituting $p = \frac{a + b + c}{2}$, we get:
    \[
    \pi \frac{(b + c - a)(c + a - b)(a + b - c) (a^2 + b^2 + c^2)}{(a + b + c)^3}
    \]

The final answer is $\boxed{\frac{\pi (a^{2} + b^{2} + c^{2})(b + c - a)(c + a - b)(a + b - c)}{(a + b + c)^{3}}}$