Problem 3. Determine the real numbers $a$ and $b$ that satisfy the equation

$$
\sqrt{3} \cdot \sqrt{a-2014}+\sqrt{b+2014}-2-\frac{a+b}{2}=0
$$

Iuliana Traşcă, Scorniceşti
## Solution and Grading Criteria

Using the inequality of means, we have:

$$
\begin{aligned}
& \sqrt{3} \cdot \sqrt{a-2014}=\sqrt{3 \cdot(a-2014)} \leq \frac{3+a-2014}{2}=\frac{a-2011}{2} \ldots \ldots \ldots \ldots \ldots . . \\
& \sqrt{b+2014}=\sqrt{1 \cdot(b+2014)} \leq \frac{1+b+2014}{2}=\frac{b+2015}{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
& \text { Then } \sqrt{3} \cdot \sqrt{a-2014}+\sqrt{b+2014} \leq \frac{a-2011+b+2015}{2}=2+\frac{a+b}{2}
\end{aligned}
$$

The relation from the statement corresponds to the equality cases in the inequalities above, so \(a-2014=3\) and \(b+2014=1\), from which \(a=2017\) and \(b=-2013\)