# Area Inequality in Triangle II

Let $ABC$ be a triangle with area $1$ and $P$ a point inside the triangle. Denote as $D,$ $E,$ $F$ the intersection points of $AP,$ $BP,$ $CP$ with the sides $BC,$ $CA,$ $AB,$ respectively.

Prove that there are at least two triangle among $PBD,$ $PCD,$ $PCE,$ $PAE,$ $PAF$ and $PBF$ with the area at most $\displaystyle\frac{1}{6}.$

### Proof

The proof has two beginnings and one end. Let $[X]$ stand for the area of of figure $X.$ Let denote by $a,$ $b,$ $c$ $[\Delta PBD],$ $[\Delta PCE],$ $[\Delta PAF],$ respectively, and by $x,$ $y,$ $z$ $[\Delta PCD],$ $[\Delta PAE],$ $[\Delta PBF].$ Let $\alpha,$ $\beta,$ $\gamma$ denote the angles around $P$ as shown below:

Then by the sine formula for the area of a triangle,

\begin{align} a &= \frac{1}{2}PB\cdot PD\sin\alpha,\\ b &= \frac{1}{2}PC\cdot PE\sin\gamma,\\ c &= \frac{1}{2}PA\cdot PF\sin\beta, \end{align}

so that $abc=PA\cdot PB\cdot PC\cdot PD\cdot PE\cdot PF\sin\alpha\sin\beta\sin\gamma.$ Clearly, $xyz$ equals the same quantity, implying $abc=xyz.$

The latter identity can be also derived from Ceva's theorem. Indeed, $\displaystyle\frac{a}{x}=\frac{DB}{DC},$ $\displaystyle\frac{b}{y}=\frac{EC}{EA},$ and $\displaystyle\frac{c}{z}=\frac{FA}{FB}.$ But, according to Ceva's theorem, $\displaystyle\frac{DB}{DC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}=1.$ So, too, $\displaystyle\frac{abc}{xyz}=1.$

Since $a+b+c+x+y+z=1,$ by the Arithmetic Mean-Geometric Mean inequality, $abc\cdot xyz\le \left(\frac{1}{6}\right)^{6}.$ Since $abc=xyz,$ both do not exceed $\left(\frac{1}{6}\right)^{3},$ implying that at least one among $a,b,c$ and at least one among $x,y,z$ does not exceed $\displaystyle\frac{1}{6}.$

### Acknowledgment

The statement above has been communicated to me by Leo Giugiuc, with credits to Professor Cristinel Mortici, Valahia University. The proof is by Leo Giugiuc.