\( \frac{1}{x}+\frac{x}{x+1} \geq 1 \) ?

To solve the inequality \( \frac{1}{x} + \frac{x}{x+1} \geq 1 \), we first find a common denominator and rewrite the expression. The common denominator is \( x(x+1) \). Therefore, we have: \[ \frac{(x+1) + x^2}{x(x+1)} \geq 1 \] This simplifies to: \[ \frac{x^2 + x + 1}{x(x+1)} \geq 1 \] Next, we can rearrange it: \[ x^2 + x + 1 \geq x(x + 1) \] Which simplifies to: \[ x^2 + x + 1 \geq x^2 + x \] Subtracting \( x^2 + x \) from both sides gives: \[ 1 \geq 0 \] This inequality holds for all \( x \) where the expressions are defined. We need to check the domain of the original expression. The terms \( \frac{1}{x} \) and \( \frac{x}{x+1} \) are defined for \( x \neq 0 \) and \( x \neq -1 \). Combining these observations, the solution to the inequality is: All \( x \) except \( x = 0 \) and \( x = -1 \), or in interval notation: \( (-\infty, -1) \cup (-1, 0) \cup (0, \infty) \).