shortlistBMO 2011

Let $D$, $E$, $F$ be the points where the incircle touches the sides $BC$, $CA$, $AB$, respectively, and let $x=AE=AF$, $y=BF=BD$, $z=CD=CE$. Express all the lengths involved in the required inequality in terms of $x$, $y$ and $z$. Clearly, $AB=x+y$, $BC=y+z$, and $CA=z+x$. To express $A{A}_1$ and $A{A}_2$, use the similarity of the triangles $A{A}_1{A}_2$ and $ABC$. Their perimeters are $2x$ and $2(x+y+z)$, respectively, so $A{A}_1/AB=A{A}_2/AC=x/(x+y+z)$, whence $A{A}_1=x(x+y)/(x+y+z)$ and $A{A}_2=x(x+z)/(x+y+z)$. Similarly, $B{B}_1=y(y+z)/(x+y+z)$, $B{B}_2=y(y+x)/(x+y+z)$, $C{C}_1=z(z+x)/(x+y+z)$ and $C{C}_2=z(z+y)/(x+y+z)$. We must show that $$9\sum x^2(x+y)(x+z)\ge (x+y+z{)}^2\sum (x+y{)}^2.$$ Alternatively, but equivalently, $$9\sum x^4+3\left(\sum x^2\right)\left(\sum xy\right)\ge 2{\left(\sum x^2\right)}^2+4{\left(\sum xy\right)}^2,$$ which is a consequence of the Cauchy-Schwarz inequality: $$3(x^4+y^4+z^4)\ge (x^2+y^2+z^2{)}^2$$ and $$x^2+y^2+z^2\ge xy+yz+zx.$$ Clearly, equality holds if and only if $x=y=z$; that is, if and only if the triangle $ABC$ is equilateral.