2009 Zhautykov International Olympiad Problems/Problem 3

First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices.

Let the incircle of $\triangle ABC$ touch $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ respectively. By Ceva's theorem, it is easy to see that $AD$ , $BE$ , $CF$ are concurrent at a point $Q$ , which we call the Gergonne point of $\triangle ABC$ .

Let $AE=AF=x$ , $BF=BD=y$ , $CD=CE=z$ . As shown in the figure, by Stewart's theorem, we have $AD^2=\frac{(z+x)^2y+(x+y)^2z}{y+z}-yz=\frac{x(4yz+zx+xy)}{y+z},$ In $\triangle ACD$ , by Menelaus' theorem, we have $\frac{AQ}{AD-AQ}\cdot \frac y{y+z}\cdot \frac zx=1,$ Solving this, we get $AQ=\frac{x(y+z)}{yz+zx+xy}\cdot AD=\frac{x\sqrt{x(y+z)(4yz+zx+xy)}}{yz+zx+xy}.$

Noting that by AM-GM inequality, we have \begin{align*} \sqrt{x(y+z)(4yz+zx+xy)}&=\frac1{\sqrt3}\sqrt{3x(y+z)(4yz+zx+xy)} \\ & \leqslant \frac{3x(y+z)+4yz+zx+xy}{2\sqrt3} \\ & =\frac{2(yz+zx+xy)}{\sqrt3}, \end{align*} we get $AQ\leqslant \frac2{\sqrt3}x,$ Thus, we obtain the following lemma.

Lemma: Let $Q$ be the Gergonne point of $\triangle ABC$ , then $AB+AC-BC\geqslant \sqrt3AQ.$

Returning to the original problem, let $K$ be the Gergonne point of $\triangle BDF$ . As shown in the figure, by the lemma, we have $AC\cdot(BD + BF - DF)\geqslant AC\cdot\sqrt3BK\geqslant 2\sqrt3S_{ABCK},$ Similarly, we have $CE\cdot(BD + DF - BF)\geqslant 2\sqrt3S_{CDEK}$ , $AE\cdot(BF + DF - BD)\geqslant 2\sqrt3S_{EFAK}$ , adding these together, we get the desired result.

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2009 Zhautykov International Olympiad Problems/Problem 3