Inequalities + gif

Let $\lfloor x-2 \rfloor =a$ .

So the given inequality becomes :

$(y^2 -y +1) \cdot a < (y^2 + y + 1)$

$\implies (a-1)y^2-(a+1)y+a-1<0$

Now we have to find the range of $a$ such that the above equation is satisfied by at least one value of $y$ .

We'll find the range of $a$ so that no value of $y$ satisfies the above inequality and then we'll remove those values from the set of $\mathbb{R}$ .

$(a-1)y^2-(a+1)y+a-1 \leq 0$

$\implies a-1>0 \ and \ (a+1)^2-4(a-1)^2 \leq 0$

$\implies a>1 \ and \ (3a-1)(a-3) \geq 0$

$\implies a>1 and \ \left(a \leq \frac{1}{3}\right) \cup \left( a \geq 3 \right)$

$\implies a \geq 3$

So for at least one value of $y$ to satisfy the given inequality :

$a \in R -[3,\infty] \implies a \in (-\infty,3)$

$\implies \lfloor x-2 \rfloor \in (-\infty,3)$

$\implies x \in (-\infty,5)$

enjoy !