I have solved the following double inequality $1\leq x < x^{2}$ as follows:
$1\leq x$ is already solved
$x< x^{2}$ using a table of sings gives me $(-\infty,0) \cup (1,\infty)$
together the solution is $(1,\infty)$.
Now my question: is the following argument correct?
$1\leq x$ is already solved
For the second inequality $x< x^{2}$ I use from the first one that $1\leq x$ so I can divide both sides of the inequality (I divide by a number different from zero and a positive number which preserves the inequality) and I get $1 < x$.
Again together the solution is $(1,\infty)$.
It is valid to use the first inequality as a condition to solve the second one?