Answer
The inequality is valid for values only less than 2 (not including it) i.e. $(-\infty,2)$.
Work Step by Step
First, we are going to move everything to the left side and simplify:
$3(x^2-2)<2(x-1)^2+x^2$
$3x^2-6<2(x^2-2x+1)+x^2$
$3x^2-6-x^2<2x^2-4x+2$
$2x^2-6-(2x^2-4x+2)<0$
$4x-8<0$
Now, we find critical points by equating the numerator and denominator to zero:
$4x-8=0$
There is only one critical point:
$4x-8=0\rightarrow 4x=8\rightarrow x=2$
Next, we are going to take two values: one less than 2 and one more than 2 to test in the original equation and check if the inequality is true or not:
First test with a value less than 2:
$3(0^2-2)<2(0-1)^2+0^2$
$3(-2)<2(-1)^2$
$-6<2(1)$
$-6<2 \rightarrow \text{ TRUE}$
Second test with a value more than 2:
$3(3^2-2)<2(3-1)^2+3^2$
$3(9-2)<2(2)^2+9$
$3(7)<2(4)+9$
$21<8+9$
$21<17 \rightarrow \text{ FALSE}$
These tests show that the inequality $3(x^2-2)<2(x-1)^2+x^2$ is valid for values only less than 2 (not including it) i.e. $(-\infty,2)$
