Solve x^2+x+1geq(x+1)(x-1)

$$x ^ { 2 } + x + 1 \geq ( x + 1 ) ( x - 1 )$$

Solve for x

$$x\geq -2$$

Solution Steps

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}+x+1\geq x^{2}-1$$

Subtract x^{2} from both sides.

$$x^{2}+x+1-x^{2}\geq -1$$

Combine x^{2} and -x^{2} to get 0.

$$x+1\geq -1$$

Subtract 1 from both sides.

$$x\geq -1-1$$

Subtract 1 from -1 to get -2.

$$x\geq -2$$