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<(x^2-x+1) < 1
You have already solved 0<(x^2-x+1) < 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 (x^2-x+1)=1 , x=0,1.
Since 1^n is always 1, this satisfies the inequation.
3) When x satisfies 1<(x^2-x+1) , 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.
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