Математикадан облыстық олимпиада, 2003-2004 оқу жылы, 11 сынып

Қосындысы 1-ге тең кез келген оң нақты $x,y$ және $z$ сандары үшін теңсіздікті дәлелде: $$\sqrt{xy+z}+\sqrt{yz+x}+\sqrt{zx+y}\le \dfrac{xy+z}{x+y}+\dfrac{yz+x}{y+z}+\dfrac{zx+y}{z+x}.$$

Комментарий/решение

$$xy+z=xy+z\cdot1=xy+z(x+y+z)=xy+zx+zy+z^2=x(y+z)+z(y+z)=(y+z)(x+z)$$

$$ yz+x=yz+x\cdot1=yz+x(z+y+x)=yz+xz+xy+x^2=z(y+x)+x(y+x)=(y+x)(z+x)$$

$$ zx+y=zx+y\cdot1=zx+y(x+z+y)=zx+yx+yz+y^2=x(z+y)+y(z+y)=(z+y)(x+y)$$

$$\sqrt{(y+z)(x+z)} +\sqrt{(y+x)(z+x)}+\sqrt{(z+y)(x+y)}\leq \frac{(y+z)(x+z)}{x+y}+\frac{(y+x)(x+z)}{z+y}+\frac{(y+z)(x+y)}{x+z}$$

$$2\sqrt{(y+z)(x+z)} +2\sqrt{(y+x)(z+x)}+2\sqrt{(z+y)(x+y)}\leq \frac{2(y+z)(x+z)}{x+y}+\frac{2(y+x)(x+z)}{z+y}+\frac{2(y+z)(x+y)}{x+z}$$

$$\left( \sqrt{\frac{(y+z)(x+z)}{x+y}}-\sqrt{\frac{(y+z)(x+y)}{x+z}}\right)^2+\left( \sqrt{\frac{(y+z)(x+y)}{x+z}}-\sqrt{\frac{(x+z)(x+y)}{y+z}}\right)^2+$$

$$+ \left( \sqrt{\frac{(y+x)(x+z)}{z+y}}-\sqrt{\frac{(y+z)(x+z)}{x+y}}\right)^2\geq 0$$

$$\sqrt{\frac{(y+z)(x+z)}{x+y}}=\alpha, \quad \sqrt{\frac{(y+z)(x+y)}{x+z}}=\beta,\quad \sqrt{\frac{(y+x)(x+z)}{z+y}}=\gamma\Rightarrow$$

$$\Rightarrow (\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2\geq 0$$