Solve 1/1+1{1+frac{1/x}}leq1

$$\frac { 1 } { 1 + \frac { 1 } { 1 + \frac { 1 } { x } } } \leq 1$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(-1,-\frac{1}{2}\right)\cup \left(0,\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$\frac{1}{1+\frac{1}{\frac{x}{x}+\frac{1}{x}}}\leq 1$$

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

$$\frac{1}{1+\frac{1}{\frac{x+1}{x}}}\leq 1$$

Variable x cannot be equal to 0 since division by zero is not defined. Divide 1 by \frac{x+1}{x} by multiplying 1 by the reciprocal of \frac{x+1}{x}.

$$\frac{1}{1+\frac{x}{x+1}}\leq 1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+1}{x+1}.

$$\frac{1}{\frac{x+1}{x+1}+\frac{x}{x+1}}\leq 1$$

Since \frac{x+1}{x+1} and \frac{x}{x+1} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{x+1+x}{x+1}}\leq 1$$

Combine like terms in x+1+x.

$$\frac{1}{\frac{2x+1}{x+1}}\leq 1$$

Variable x cannot be equal to -1 since division by zero is not defined. Divide 1 by \frac{2x+1}{x+1} by multiplying 1 by the reciprocal of \frac{2x+1}{x+1}.

$$\frac{x+1}{2x+1}\leq 1$$

Denominator 2x+1 cannot be zero since division by zero is not defined. There are two cases.

$$2x+1>0 2x+1<0$$

Consider the case when 2x+1 is positive. Move 1 to the right hand side.

$$2x>-1$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x>-\frac{1}{2}$$

The initial inequality does not change the direction when multiplied by 2x+1 for 2x+1>0.

$$x+1\leq 2x+1$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$x-2x\leq -1+1$$

Combine like terms.

$$-x\leq 0$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\geq 0$$

Now consider the case when 2x+1 is negative. Move 1 to the right hand side.

$$2x<-1$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x<-\frac{1}{2}$$

The initial inequality changes the direction when multiplied by 2x+1 for 2x+1<0.

$$x+1\geq 2x+1$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$x-2x\geq -1+1$$

Combine like terms.

$$-x\geq 0$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\leq 0$$

Consider condition x<-\frac{1}{2} specified above.

$$x<-\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-\frac{1}{2})\cup [0,\infty)$$