Bulgaria

Let $D,E$ and $F$ be the common points of with the sides $BC,CA$ and $AB$, respectively, and let $AE=AF=x$, $BF=BD=y$ and $CD=CE=z$. Since $△A{A}_1{A}_2\sim △ABC$, then $\frac{A{A}_1}{AB}=\frac{A{A}_2}{AC}=\frac{P_{A{A}_1{A}_2}}{P_{ABC}}=\frac{x}{x+y+z}$ and hence $A{A}_1=\frac{x(x+y)}{x+y+z}$ and $A{A}_2=\frac{x(x+z)}{x+y+z}$. Analogously, $B{B}_1=\frac{y(y+z)}{x+y+z}$, $B{B}_2=\frac{y(y+x)}{x+y+z}$, $C{C}_1=\frac{z(z+x)}{x+y+z}$ and $C{C}_2=\frac{z(z+y)}{x+y+z}$. Then the given inequality is equivalent to $$9\sum x^2(x+y)(x+z)\ge (x+y+z{)}^2\sum (x+y{)}^2,$$ i.e. $$9\sum x^4+2(\sum x^2)(\sum xy)\ge 2(\sum x^2{)}^2+4(\sum xy{)}^2,$$

The last inequality follows by the well-known inequalities $$3(x^4+y^4+z^4)\ge (x^2+y^2+z^2{)}^2\text{ and }{x}^2+y^2+z^2\ge xy+yz+zx.$$