[ Quadratic equations and systems of equations ] [Completing the square. Sums of squares] Solve the system: $\left\{\begin{array}{c}x-y \geq z \\ x^{2}+4 y^{2}+5=4 z\end{array}\right.$.
Multiply both sides of the inequality by 4 and substitute the value of $4z$ from the equation. We get:

$x^{2}+4 y^{2}+5 \leq 4 x-4 y \Leftrightarrow(x-2)^{2}+(2 y+1)^{2} \leq 0 \Leftrightarrow x=2, y=-0.5$. Substituting the found values into the equation, we get $z=2.5$.

## Answer

$(2; -0.5; 2.5)$.