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!
 A: You can't divide by $x+3$ if $x=-3$. Furthermore, if $x<-3$, then $x+3<0$. So, after dividing by $x+3$, you will get $x-3\geqslant x$ in this case (which is impossible, of course).
A: 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 $x \geq -3.$
A: Dividing both sides of the inequality
$$
(x-3)(x+3) \leq x(x+3)
$$
by $x+3$ is problematic for two reasons. The first is that it can lead to division by zero, as you have already noted. The second is that when you multiply or divide an inequality by a negative number, you have to switch the direction in which the inequality is written. Hence, if $x+3$ is smaller than $0$, then you have to flip the $\leq$ to $\geq$. If you insist upon dividing by $x+3$, this approach can be made to work if you split the problem into cases.
Case 1: if $x+3>0$, then dividing both sides by $x+3$ does not switch the direction of the inequality. Hence,
$$
(x-3) \leq x \\
-3 \leq 0 \, .
$$
The final statement is true, and so for $x+3>0$, the inequality must hold.
Case 2: if $x+3=0$, then we can't divide both sides by $x+3$. But notice that if $x+3=0$, then
$$
(x-3)(0) \leq x(0)
$$
which is the same as $0 \leq 0$. Hence, the inequality is true for $x+3=0$.
Case 3: if $x+3<0$, then we have to switch the direction of the inequality, and we get that
$$
x-3 \geq x \\
-3 \geq 0
$$
The above inequality has no solutions, and so the inequality is not true for $x+3<0$. Putting everything together, we find that $x+3$ must be greater than or equal to $0$ for the inequality to be true. The solution is thus $x\geq -3$.

In general, if $a \geq b$, then $-a \leq -b$. Hence, if we multiply both sides by $-1$, we have to switch the direction of the inequality. This is because
\begin{align}
& a \geq b \\
\implies & 0 \geq b - a \\
\implies & -b \geq -a \\
\implies & -a \leq -b \, .
\end{align}
