Maths Olympiad Prep

Library / /1 of 33

Geometry Difficulty 7.2 National olympiad, round 2 Prove it Turkey

Let xx, yy, zz be positive real numbers such that
x,y,z are sides of a triangle and xy+yz+zx=5. \sqrt{x}, \sqrt{y}, \sqrt{z} \text{ are sides of a triangle and } \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 5.
Prove that
x(y2−2z2)z+y(z2−2x2)x+z(x2−2y2)y≥0. \frac{x(y^2 - 2z^2)}{z} + \frac{y(z^2 - 2x^2)}{x} + \frac{z(x^2 - 2y^2)}{y} \ge 0.

Solution

Since x\sqrt{x}, y\sqrt{y}, z\sqrt{z} are sides of a triangle, we have
(x+y+z)(x+y−z)(y+z−x)(z+x−y)=2(xy+yz+zx)−x2−y2−z2≥0. (\sqrt{x} + \sqrt{y} + \sqrt{z})(\sqrt{x} + \sqrt{y} - \sqrt{z})(\sqrt{y} + \sqrt{z} - \sqrt{x})(\sqrt{z} + \sqrt{x} - \sqrt{y}) = 2(xy + yz + zx) - x^2 - y^2 - z^2 \ge 0.
We also have
5xy=(xy+yz+zx)xy=x2+xy2z+yz 5xy = \left( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \right) xy = x^2 + \frac{xy^2}{z} + yz
5yz=(xy+yz+zx)yz=y2+yz2x+zx 5yz = \left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x}\right) yz = y^2 + \frac{yz^2}{x} + zx
5zx=(xy+yz+zx)zx=z2+zx2y+xy 5zx = \left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x}\right) zx = z^2 + \frac{zx^2}{y} + xy
and by summing up the equations above, we get
2(xy+yz+zx)−(x2+y2+z2)=xy2z+yz2x+zx2y−2(xy+yz+zx) 2(xy + yz + zx) - (x^2 + y^2 + z^2) = \frac{x y^2}{z} + \frac{y z^2}{x} + \frac{z x^2}{y} - 2(xy + yz + zx)
=x(y2−2z2)z+y(z2−2x2)x+z(x2−2y2)y≥0. = \frac{x(y^2 - 2z^2)}{z} + \frac{y(z^2 - 2x^2)}{x} + \frac{z(x^2 - 2y^2)}{y} \geq 0.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.