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Problem A. 405. (September 2006)

A. 405. The real numbers a, b, c, x, y, z satisfy a\geb\gec>0 and x\gey\gez>0. Prove that

\frac{a^2x^2}{(by+cz)(bz+cy)} + \frac{b^2y^2}{(cz+ax)(cx+az)} + \frac{c^2z^2}{(ax+by)(ay+bx)} \ge \frac34.

(Korean competition problem)

(5 pont)

Deadline expired on October 16, 2006.

Solution. We will use Nesbitt's inequality: for arbitrary positive numbers A,B,C,

\frac{A}{B+C}+\frac{B}{C+A}+\frac{C}{A+B}\ge\frac32.

(This is equivalent to the AM-HM inequality for numbers B+C, C+A and A+B.)

By the rearrangement inequality, bz+cy\leby+cz and

(by+cz)(bz+cy)\le(by+cz)2\le2(b2y2+c2z2).

It can be obtained similarly that

(cz+ax)(cx+az)\le2(c2z2+a2x2)

and

(ax+by)(ay+bx)\le2(a2x2+b2y2).

Applying these estiamtes and Nesbitt's inequality on the numbers a2x2, b2y2 and c2z2,

\frac{a^2x^2}{(by+cz)(bz+cy)} + \frac{b^2y^2}{(cz+ax)(cx+az)} + \frac{c^2z^2}{(ax+by)(ay+bx)} \ge

\ge \frac12\left( \frac{a^2x^2}{b^2y^2+c^2z^2} + \frac{b^2y^2}{c^2z^2+a^2x^2} + \frac{c^2z^2}{a^2x^2+b^2y^2} + \right) \ge \frac12\cdot\frac32 = \frac34.

Statistics:

26 students sent a solution.
5 points:Blázsik Zoltán, Dobribán Edgár, Farkas Ádám László, Fischer Richárd, Gyenizse Gergő, Gyürke Csaba, Hujter Bálint, Károlyi Márton, Kisfaludi-Bak Sándor, Kónya 495 Gábor, Korándi Dániel, Kornis Kristóf, Kutas Péter, Lovász László Miklós, Nagy 224 Csaba, Nagy 235 János, Nagy 314 Dániel, Sümegi Károly, Szilágyi Dániel, Szirmai Péter, Szűcs 003 Gábor, Tomon István, Tossenberger Anna, Varga 171 László.
1 point:1 student.
0 point:1 student.

Problems in Mathematics of KöMaL, September 2006