$x^2 + 4x + y^2 - y + 15$
$= (x^2 + 4x + 4) + \left(y^2 - 2\cdot y \cdot\dfrac12+\dfrac14\right) + \dfrac{43}{4}$
$= (x+2)^2 +\left(y -\dfrac12\right)^2 +\dfrac{43}{4}$
Ta có:
$\begin{cases}(x+2)^2 \quad \forall x\\\left(y -\dfrac12\right)^2 \geq 0\quad \forall y\end{cases}$
$\to (x+2)^2 +\left(y -\dfrac12\right)^2 +\dfrac{43}{4} \geq \dfrac{43}{4} > 0$
$\to (x+2)^2 +\left(y -\dfrac12\right)^2 +\dfrac{43}{4} > 0$
Vậy $x^2 + 4x + y^2 - y + 15>0\quad \forall x;y$
