Division in inequality I have the problem
$(x-3)*(x+3) \leq x*(3+x)$
and with the following steps
$(x-3)*(x+3) \leq x*(3+x)$
$x^2-9 \leq 3x+x^2$
$-9 \leq 3x$
$-3 \leq x$.
However, my first thought was not to go this way but rather divide by $(3+x)$, which yields
$x-3 \leq x$.
Obviously this does not work. My first thought was that it is not allowed to divide by $(3+x)$ since this contains the solution -3 for x, which would equal to 0 $(-3+3) = 0$.
Question: Why does this not work? Is my thought (it does not work since it would include a division with 0) correct?
Thank you for your help!

Actually I think you can solve this problem by dividing by $x+3,$ provided that you handle the sign of this term correctly and also take care of the special case
$x + 3 = 0.$
Consider three cases: $x + 3 = 0,$ $x + 3 > 0$, $x + 3 < 0.$
Case $x + 3 = 0$:
In this case $(x-3)(x+3) = 0 = x(3+x),$ so it is certainly the case that
$(x-3)(x+3) \leq x(3+x)$. In this case $x = -3,$ so $x = -3$ is one solution.
Case $x + 3 > 0$:
In this case, multiplying or dividing by $x + 3$ on each side of an inequality leaves the inequality intact, that is, $(x-3)(x+3) \leq x(3+x)$ if and only if $x-3 \leq x$.
But $x-3 \leq x$ for all $x$, so every $x$ such that $x + 3 > 0$ is a solution.
In other words, every $x > -3$ is a solution.
Case $x + 3 < 0$:
In this case, multiplying or dividing by $x + 3$ on each side of an inequality reverses the direction of the inequality, that is, $(x-3)(x+3) \leq x(3+x)$ if and only if $x-3 \geq x$.
But there is no $x$ such that $x-3 \geq x$ for all $x$, so this case does not find any solutions.
Conclusion: The only solutions are the ones found in the first two cases, namely, $x$ must satisfy either $x = -3$ or $x > -3.$ A more succinct way of expressing this is that $x \geq -3.$
