To solve the inequality [tex]\(\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+2}\)[/tex], follow these steps:
1. Rewrite the inequality with a common denominator:
First, rewrite the inequality to combine the fractions on the left side. The common denominator for the fractions on the left side is [tex]\( x(x+1) \)[/tex], and for the right side, it is [tex]\( x(x+1)(x+2) \)[/tex]:
[tex]\[
\frac{(x+1) + x}{x(x+1)} < \frac{2}{x+2}
\][/tex]
Simplify the numerator on the left side:
[tex]\[
\frac{2x+1}{x(x+1)} < \frac{2}{x+2}
\][/tex]
2. Cross-multiply to eliminate the fractions:
To eliminate the fractions, cross-multiply (assuming [tex]\( x \)[/tex], [tex]\( x+1 \)[/tex], and [tex]\( x+2 \)[/tex] are non-zero):
[tex]\[
(2x+1)(x+2) < 2x(x+1)
\][/tex]
3. Expand both sides of the inequality:
Expand the products:
[tex]\[
2x^2 + 5x + 2 < 2x^2 + 2x
\][/tex]
4. Simplify the inequality:
Subtract [tex]\( 2x^2 + 2x \)[/tex] from both sides:
[tex]\[
5x + 2 - 2x < 0
\][/tex]
[tex]\[
3x + 2 < 0
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
Isolate [tex]\( x \)[/tex]:
[tex]\[
3x < -2
\][/tex]
[tex]\[
x < -\frac{2}{3}
\][/tex]
6. Consider the nonlinearity and potential critical points:
The solution also needs to consider the points where either the denominator of the fractions becomes zero. These critical points are [tex]\( x = 0 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = -2 \)[/tex]. So, we need to analyze the intervals determined by these points:
- [tex]\( (-\infty, -2) \)[/tex]
- [tex]\( (-2, -1) \)[/tex]
- [tex]\( (-1, 0) \)[/tex]
- [tex]\( (0, \infty) \)[/tex]
Evaluating each interval by testing points within them:
- For [tex]\( x \in (-\infty, -2) \)[/tex]:
It does not satisfy the inequality.
- For [tex]\( x \in (-2, -1) \)[/tex]:
Test with [tex]\( x = -1.5 \)[/tex]:
[tex]\[
\frac{1}{-1.5} + \frac{1}{-0.5} < \frac{2}{0.5}
\][/tex]
[tex]\[
-\frac{2}{3} - 2 < 4
\][/tex]
True.
- For [tex]\( x \in (-1, 0) \)[/tex]:
Test with [tex]\( x = -0.5 \)[/tex]:
[tex]\[
\frac{1}{-0.5} + \frac{1}{0.5} < \frac{2}{1.5}
\][/tex]
[tex]\[
-2 + 2 < \frac{4}{3}
\][/tex]
True.
Thus, combining these results:
[tex]\[
x \in (-2, -1) \cup \left(-\frac{2}{3}, 0\right)
\][/tex]
Therefore, the solution set for the inequality [tex]\(\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+2}\)[/tex] is:
[tex]\[
\boxed{(-2, -1) \cup \left(-\frac{2}{3}, 0\right)}
\][/tex]
