Prove that $\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$ is a closed and bounded set. 
Prove that the following set $$V=\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$$ is closed and bounded.

The set $V$ is closed. In fact, for any continuous function $f:\mathbb R^n\rightarrow \mathbb R$, the set $\{\textbf{x}\in\mathbb R^n: f(\textbf{x})=0\}$ is closed. I would need to have a hint for the boundedness of $V$. Thanks!

Note that $(a-b)^2 \ge 0$, so $ab \le \frac{a^2+b^2}{2}$. Thus, since $x^2 - xy + y^2 - 1 = 0$ we have
$x^2 + y^2 = 1 + xy \le 1 + \frac{x^2+y^2}{2}$
and so
$\frac{1}{2} (x^2 + y^2) \le 1$
and finally
$x^2 + y^2  \le 2$.
