[ $\underline{\text { Quadratic Inequalities and Systems of Inequalities }}$ ]

For positive numbers $x$ and $y$, they compute $a=1 / y$ and $b=y+1 / x$. After that, they find $C$ - the smallest number among the three: $x, a$, and $b$.

What is the largest value that $C$ can take?
From the condition, it follows that $C \leq x, C \leq a$ and $C \leq b$. Since $b=1 / a+1 / x$, then $C \leq \frac{1 / a+1 / x}{}$. Moreover, since all numbers are positive, then ${ }^{1} / a \leq 1 / C$ and

$1 / x \leq 1 / C$. Therefore, $C \leq \frac{1}{} / C+\frac{1}{} / C$, which means $C \leq \sqrt{2}$.

The value $\sqrt{2}$ is achieved if $x=\sqrt{2}, y=\frac{1}{\sqrt{2}}$, since in this case $a=\sqrt{2}$ and $b=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$.

## Answer

$\sqrt{2}$.