`(4x)/(x^(2) +3)ge 1.`

Answer

To solve the inequality \(\frac{4x}{x^2 + 3} \geq 1\), we will follow these steps:

### Step 1: Rearranging the Inequality
First, we rearrange the inequality to bring all terms to one side:
\[ \frac{4x}{x^2 + 3} - 1 \geq 0 \]
This can be rewritten as:
\[ \frac{4x - (x^2 + 3)}{x^2 + 3} \geq 0 \]
Simplifying the numerator gives us:
\[ \frac{4x - x^2 - 3}{x^2 + 3} \geq 0 \]
Thus, we have:
\[ \frac{-x^2 + 4x - 3}{x^2 + 3} \geq 0 \]

### Step 2: Factoring the Numerator
Next, we factor the numerator \(-x^2 + 4x - 3\). We can rewrite it as:
\[ -(x^2 - 4x + 3) \]
Now, we factor \(x^2 - 4x + 3\):
\[ x^2 - 4x + 3 = (x - 3)(x - 1) \]
So, we have:
\[ \frac{-(x - 3)(x - 1)}{x^2 + 3} \geq 0 \]

### Step 3: Analyzing the Denominator
The denominator \(x^2 + 3\) is always positive since \(x^2\) is non-negative for all real \(x\) and adding 3 ensures it is positive. Therefore, we only need to focus on the numerator:
\[ -(x - 3)(x - 1) \geq 0 \]

### Step 4: Finding Critical Points
The critical points from the numerator are \(x = 1\) and \(x = 3\). We will analyze the sign of the expression in the intervals determined by these points: \((-\infty, 1)\), \((1, 3)\), and \((3, \infty)\).

### Step 5: Testing Intervals
1. **Interval \((-\infty, 1)\)**: Choose \(x = 0\):
\[ -(0 - 3)(0 - 1) = -(-3)(-1) = -3 < 0 \]

2. **Interval \((1, 3)\)**: Choose \(x = 2\):
\[ -(2 - 3)(2 - 1) = -(-1)(1) = 1 > 0 \]

3. **Interval \((3, \infty)\)**: Choose \(x = 4\):
\[ -(4 - 3)(4 - 1) = -1(3) = -3 < 0 \]

### Step 6: Conclusion
The expression \(- (x - 3)(x - 1) \geq 0\) is satisfied in the interval \([1, 3]\). Therefore, the solution to the inequality \(\frac{4x}{x^2 + 3} \geq 1\) is:
\[ x \in [1, 3] \]