Solve the inequation `(x^(2)+x+1)^(x)lt1`

Explanation

Taking logarithm both sides on the base 10, then `xlog(x^(2)+x+1)lt0` brgt which is equivalent to the collection of systems
`[({(xgt0),(log(x^(2)+x+1)lt0):}),({(xlt0),(log(x^(2)+x+1)gt0):}):}implies`[({(xgt0),(x^(2)+x+1lt1):}),({(xlt0),(x^(2)+x+1gt1):}):}`
`implies `[({(xgt0),(x(x+1)lt0):}),({(xlt0),(x(x+1)gt0):}):}implies`[({(xgt0),(-1ltxlt0):}),({(xlt0),(xgt0 "and"xlt(-1)):}):}`
`implies{(x epsilon phi),(x lt(-1)):}`
Consequently the interval `x epsilon(-oo,-1)` is the set of all solutions of the original inequation.