Математикадан республикалық олимпиада, 2002-2003 оқу жылы, 11 сынып

$x,y,z$ оң нақты сандары үшін келесі теңсіздікті дәлелдеңіздер: $\frac{{{x}^{3}}}{x+y}+\frac{{{y}^{3}}}{y+z}+\frac{{{z}^{3}}}{z+x}\ge \frac{xy+yz+zx}{2}.$

Комментарий/решение

$$\frac{x^3}{x+y}+\frac{y^3}{y+z}+\frac{z^3}{z+x}\geq \frac{xy+yz+zx}{2}$$

$$\frac{x^3}{x+y}+\frac{y^3}{y+z}+\frac{z^3}{z+x}+\frac{xy+yz+zx}{4}\geq \frac{3xy+3yz+3zx}{4}$$

$$\frac{x^3}{x+y}+\frac{y^3}{y+z}+\frac{z^3}{z+x}+\frac{xy+yz+zx}{4}+\frac{x^2+y^2+z^2}{4}\geq \frac{3xy+3yz+3zx}{4}+\frac{x^2+y^2+z^2}{4}$$

$$\left(\frac{x^3}{x+y}+\frac{x(x+y)}{4}\right)+\left(\frac{y^3}{z+y}+\frac{y(z+y)}{4}\right)+\left(\frac{z^3}{x+z}+\frac{z(x+z)}{4}\right) \geq \frac{3xy+3yz+3zx}{4}+\frac{x^2+y^2+z^2}{4}$$

$$\left(\frac{x^3}{x+y}+\frac{x(x+y)}{4}\right)+\left(\frac{y^3}{z+y}+\frac{y(z+y)}{4}\right)+\left(\frac{z^3}{x+z}+\frac{z(x+z)}{4}\right) \geq $$

$$\geq 2\sqrt{\frac{x^3}{x+y}\cdot\frac{x(x+y)}{4}}+2\sqrt{\frac{y^3}{z+y}\cdot\frac{y(z+y)}{4}}+2\sqrt{\frac{z^3}{x+z}\cdot\frac{z(x+z)}{4}}=x^2+y^2+z^2$$

$$ x^2+y^2+z^2\geq \frac{3xy+3yz+3zx}{4}+\frac{x^2+y^2+z^2}{4} \Rightarrow$$

$$\Rightarrow x^2+y^2+z^2\geq xy+zx+zy\Leftrightarrow (x-z)^2+(z-y)^2+(y-x)^2\geq0$$