# Question e8275

Sep 28, 2017

The problem is that you had missed case analysis.

#### Explanation:

Suppose $a$ to be a positive value except for $1$.
If $a$ is larger than $1$, the following is true.

i) If $1 < a$, ${a}^{x} < {a}^{y}$ if and only if $x < y$.

However, if $a$ is smaller than 1, the magnitude reverces.
ii) If $0 < a < 1$, ${a}^{x} < {a}^{y}$ if and only if $x > y$.

Therefore, you must solve the inequation as follows:

1) When $x$ satisfies $0 < \left({x}^{2} - x + 1\right) < 1$
You have already solved $0 < \left({x}^{2} - x + 1\right) < 1$ and got $0 < x < 1$.
This time,
(x^2-x+1)^-1 ≦ (x^2-x+1)^x  ⇔　-1 ≧ x

The common part of $0 < x < 1$ and -1 ≧ x is null.

2) When $x$ satisfies $\left({x}^{2} - x + 1\right) = 1$, $x = 0 , 1$.
Since ${1}^{n}$ is always 1, this satisfies the inequation.
3) When $x$ satisfies $1 < \left({x}^{2} - x + 1\right)$, the range of $x$ is
$x < 0$ or $1 < x$.

At that time,
(x^2-x+1)^-1 ≦ (x^2-x+1)^x  ⇔　-1 ≦ x
and the common part of ($x < 0$ or $1 < x$) and $- 1 < x$
is -1≦ x < 0 or $1 < x$.

Now you got to the answer -1≦ x ≦ 0 or 1≦ x#.

The answer was checked at http://www.wolframalpha.com