2010 JMO #4

4. 양의 실수 x,y,z에 대해 다음 부등식이 성립함을 보여라.

\displaystyle \sum_{cyc} \frac{1+xy+xz}{(1+y+z)^2} \geq 1

코시-슈바르츠 부등식으로 (1+xy+xz)(1+\frac{y}{x} + \frac{z}{x}) \geq (1+y+z)^2가 성립하니까
\displaystyle \sum_{cyc} \frac{1+xy+xz}{(1+y+z)^2} \geq \sum_{cyc} \frac{1}{1+\frac{y}{x}+\frac{z}{x}} = \sum_{cyc} \frac{x}{x+y+z}=1. 끝

등호는 x=y=z=1일 때 성립한다.

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