Solve : `(1)/(x^(2) +x) le(1)/(2x^(2) + 2x+3)`

Answer

To solve the inequality

\[ \frac{1}{x^2 + x} \leq \frac{1}{2x^2 + 2x + 3}, \]

we will follow these steps:

### Step 1: Rearranging the Inequality
First, we can rearrange the inequality by moving all terms to one side:

\[ \frac{1}{2x^2 + 2x + 3} - \frac{1}{x^2 + x} \geq 0. \]

### Step 2: Finding a Common Denominator
Next, we find a common denominator for the fractions:

\[ \frac{(x^2 + x) - (2x^2 + 2x + 3)}{(2x^2 + 2x + 3)(x^2 + x)} \geq 0. \]

### Step 3: Simplifying the Numerator
Now, simplify the numerator:

\[ x^2 + x - (2x^2 + 2x + 3) = x^2 + x - 2x^2 - 2x - 3 = -x^2 - x - 3. \]

So, we can rewrite the inequality as:

\[ \frac{-x^2 - x - 3}{(2x^2 + 2x + 3)(x^2 + x)} \geq 0. \]

### Step 4: Factoring the Numerator
Next, we factor the numerator. The expression \(-x^2 - x - 3\) can be rewritten as:

\[ -(x^2 + x + 3). \]

### Step 5: Analyzing the Denominator
Now, we analyze the denominator:

1. \(2x^2 + 2x + 3\) is always positive since its discriminant \(b^2 - 4ac = 2^2 - 4 \cdot 2 \cdot 3 = 4 - 24 < 0\).
2. \(x^2 + x\) is zero at \(x = 0\) and \(x = -1\), and it is positive for \(x < -1\) and \(x > 0\).

### Step 6: Finding the Roots of the Numerator
Now, we need to find the roots of the numerator \(x^2 + x + 3 = 0\):

The discriminant is \(b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 3 = 1 - 12 < 0\), which means the quadratic does not have real roots and is always positive.

### Step 7: Analyzing the Sign of the Expression
Since the numerator \(- (x^2 + x + 3)\) is always negative and the denominator is positive except at the points \(x = 0\) and \(x = -1\), we can conclude:

- The entire expression is negative for \(x < -1\) and \(x > 0\).
- The expression is undefined at \(x = 0\) and \(x = -1\).

### Step 8: Conclusion
The solution to the inequality is:

\[ x \in (-1, 0). \]