Solving $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$ $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$
So, getting rid of the denominators I got to:
$(x+1)(x+5) + x(x+5) < 2x(x+1)$
$\rightarrow x < \frac{-5}{9}$
And also keeping in mind we can't have $-1,-5$ in the solution set, I got that the solution set was:
$\left(-\infty ,-5\right)\cup \left(-5,-1\right)\cup \left(-1,-\frac{5}{9}\right)$
But this is not correct. Help appreciated!
 A: When multiply we need to take into account the sign; in this case your step are not correct since you are assuming positive terms.
As an alternative to avoid multiplication, we have
$$\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5} \iff\frac{1}{x} + \frac{1}{x+1} - \frac{2}{x+5}<0 \iff \frac{9x+5}{x(x+1)(x+5)}<0$$
then study the sign of each term.
A: $$\frac 1x + \frac 1{x+1} - \frac 2{x+5} < 0$$
$$\implies{ 9x+5\over x(x+1)(x+5)} < 0$$
Wherever the above statement is true we have our required points.
To make it easier to study, draw a rough sketch of the graph. It is not as mathematically rigorous as simply inspecting the sign, but is perhaps more intuitive. The processes are equivalent anyway and if you must you can simply use a table of critical points and the signs of the fuction between them. We have 4 critical points: $x = -5, x = -1, x = \frac{-5}9, x = 0$, where the numerator or denominator becomes zero.
Coming from the right of $x = 0$, all the factors in both the numerator and denominator are positive, so we can "draw" a very rough sketch of the graph above the line. The actual behavior of the curve is unimportant here, but if it troubles you, you may simply mark it as positive or negative. 
As we approach $0$, the denominator approaches $0$ also so the curve moves up toward $\infty$. Similarly, the denominator is negative but indeterminate as we go just left of $0$. As we approach $\frac {-5}9,$ the numerator approaches $0$ from below as does our expression. Similarly we may continue and get the result we require, which is that $x \epsilon (-5,-1) \cup (-\frac 59, 0)$
I include here the actual graph of the function:
(The function is in red. The lines $x =$ critical point are in black and are denoted by labels of various colors by the points they intersect with the $x$ axis. 

A: To clear the fractions, you need to multiply by $|x(x+1)(x+5)|$ or $[x(x+1)(x+5)]^2.$ In these cases you can be sure that the inequality is preserved, since they are never negative.
