Problem A. 388. (December 2005)

A. 388. In the hexagon ABCDEF we have AB=BC, CD=DE and EF=FA. Prove that

$\frac{AB}{BE}+\frac{CD}{DA}+\frac{EF}{FC}\ge\frac32.$

(5 pont)

Deadline expired on January 16, 2006.

Solution. Apply Ptolemaios' inequality for the quadrilateral ABCE:

AB^.CE+BC^.EA $\ge$ AC^.BE.

Knowing that AB=BC,

AB^.(CE+EA) $\ge$ AC^.BE,

$\frac{AB}{BE} \ge \frac{AC}{CE+EA}.$

Similarly

$\frac{CD}{DA} \ge \frac{CE}{EA+AC}$

and

$\frac{EF}{FC} \ge \frac{EA}{AC+CE}.$

Let AC=x, CE=y and EA=z. Summing up the inequalities above,

$\frac{AB}{BE}+\frac{CD}{DA}+\frac{EF}{FC}\ge \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}.$

By the AM-HM inequality

$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} =$

$= \left(\frac{x+y+z}{y+z}-1\right)+ \left(\frac{x+y+z}{z+x}-1\right)+ \left(\frac{x+y+z}{x+y}-1\right)=$

$= (x+y+z)\left(\frac1{y+z}+\frac1{z+x}+\frac1{x+y}\right)-3 \ge$

$\ge (x+y+z)\cdot\frac{9}{(y+z)+(z+x)+(x+y)}-3=\frac32.$

Statistics:

15 students sent a solution.

5 points: Blázsik Zoltán, Dücső Márton, Erdélyi Márton, Estélyi István, Fischer Richárd, Gyenizse Gergő, Hujter Bálint, Jankó Zsuzsanna, Kisfaludi-Bak Sándor, Kónya 495 Gábor, Kutas Péter, Nagy 224 Csaba, Paulin Roland, Tomon István, Viktor Simjanoski.

Problems in Mathematics of KöMaL, December 2005

15 students sent a solution.
5 points:	Blázsik Zoltán, Dücső Márton, Erdélyi Márton, Estélyi István, Fischer Richárd, Gyenizse Gergő, Hujter Bálint, Jankó Zsuzsanna, Kisfaludi-Bak Sándor, Kónya 495 Gábor, Kutas Péter, Nagy 224 Csaba, Paulin Roland, Tomon István, Viktor Simjanoski.

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