Problem 5. Solve the system of equations $\left\{\begin{array}{l}\sqrt{2016 \frac{1}{2}+x}+\sqrt{2016 \frac{1}{2}+y}=114 \\ \sqrt{2016 \frac{1}{2}-x}+\sqrt{2016 \frac{1}{2}-y}=56\end{array}\right.$.
# Solution:

We will use the inequality $\sqrt{\frac{a^{2}+b^{2}}{2}} \geq \frac{a+b}{2}$ (the quadratic mean of two numbers is greater than or equal to their arithmetic mean). Note that the right-hand side of the inequality equals the left-hand side when $a$ and $b$ are equal.

By taking one of the roots of the equation as $a$ and the other as $b$, we obtain the following inequalities:

$$
\left\{\begin{array}{l}
\sqrt{\frac{2016 \frac{1}{2}+x+2016 \frac{1}{2}+y}{2}} \geq \frac{114}{2} \\
\sqrt{\frac{2016 \frac{1}{2}-x+2016 \frac{1}{2}-y}{2}} \geq \frac{56}{2}
\end{array},\left\{\begin{array}{l}
\sqrt{\frac{4033+x+y}{2}} \geq \frac{114}{2} \\
\sqrt{\frac{4033-x-y}{2}} \geq \frac{56}{2}
\end{array},\left\{\begin{array}{l}
\frac{4033}{2}+\frac{x+y}{2} \geq 57^{2} \\
\frac{4033}{2}-\frac{x+y}{2} \geq 28^{2}
\end{array}\right.\right.\right.
$$

$$
\left\{\begin{array}{l}
\frac{x+y}{2} \geq 1232.5 \\
\frac{x+y}{2} \leq 1232.5
\end{array}, \text { therefore, } \frac{x+y}{2}=1232.5\right.
$$

In this case, both inequalities become equalities, so $2016 \frac{1}{2} \pm x=2016 \frac{1}{2} \pm y$, which means $x=y=1232.5$

Answer: $\left(1232 \frac{1}{2} ; 1232 \frac{1}{2}\right)$