Question

What is the domain of this inequality?
1/(x²-x+1) ≤ (x²-x+1)^x

Answer 1

\[\frac{ 1 }{ x^2-x+1 } \le (x^2-x+1)^x\] Here's a friendlier display for the eyes.

Answer 2

\[ \frac{1}{x^2-x+1} ≤ (x²-x+1)^x \]
for ease of typing, let \( y = x^2-x+1 \) and write
\[ y^{-1} ≤ y^x \]
take the log of both sides
\[ -1 \ln(y) ≤ x \ln(y) \]
add ln(y) to both sides
\[ 0 ≤ x \ln(y) + \ln(y) \\ 0 ≤ (x+1) \ln(y) \]

the expression will be positive (i.e. 0 or greater) if both factors are positive:
x+1 ≥ 0 —> x ≥ -1
and at the same time ln(y) ≥ 0 —> y ≥ 1 —> \( x^2 -x + 1 ≥ 1 \)
this latter condition requires
\[ x^2 -x ≥ 0 \\ x(x-1) ≥ 0 \]
again, both factors must be positive to get a positive result (or both factors must be negative, which we will look at next)
x ≥ 0 and x≥ 1
along with the requirement x ≥ -1 simplifies to x≥ 1

if both factors are negative, we require x≤0 and x≤ 1 i.e. x≤ 0
along with the requirement x ≥ -1 gives -1 ≤ x ≤ 0


We now consider both factors (x+1) and ln(y) being negative in
\[ 0 ≤ (x+1) \ln(y) \]
this requires x+1 ≤ 0 —> x ≤ -1
and ln(y) ≤ 0 —> y ≤ 1 —> \[ x^2 -x + 1≤ 1 \\ x(x-1) ≤0\]
this latter condition requires each factor to have opposite signs.
1) x≥0 and x≤ -1 There is no solution.
or
2) x≤0 and x ≥ -1
along with x ≤ -1. There is only the solution x=-1 which is covered by the solution up above.


In summary, the domain is:
-1 ≤ x ≤ 0
and
x≥ 1