Inequality $x^2\leq y$ I have a question about how to handle this inequality:
$$x^2\leqslant y \to x\leqslant \pm\sqrt{y}$$ or it should be $$\sqrt{x^2}\leq \sqrt{y}\Rightarrow$$ 
either $$x\leq\sqrt{y}$$
  or
  $$-x\leq\sqrt{y}\Rightarrow x\geq-\sqrt{y}$$ so$$-\sqrt{y}\leq x\leq\sqrt{y}$$
Is my way of thinking is correct?

If $y<0$ $$ x^2 \le y$$ does not have any real solution.
For $y=0$, the only solution is $x=0$
For  $ y>0$, we get   $$ x^2 \le y \iff |x|\le \sqrt y \iff  -\sqrt y \le x \le \sqrt y $$
