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?

I assume you work in $\mathbb{R}$. If $y$ is negative, then no $x$ satisfies the inequality. If $y$ is non-negative, then $-\sqrt{y} \leq x \leq \sqrt{y}$.
