Value of $y$ for inequality to hold true

Question

What are the integral values of $y$ for which $(y^2-5y+3)(x^2+x+1)<2x$ . I think graphing is the best way to solve it but this question came in my paper where there is no big place to graph $3-4$ points of each curve. So my next thought was using range of $y^2-5y+3,\frac{2x}{x^2+x+1}$. Also note that $x^2+x+1>0$ for all real $x$ . But now I cannot implement it correctly

Answer 1

Divide and conquer.

Try to solve $x^2+x+1 <2x$

$(x - \frac1{2})^2 + \frac{3}{4} < 0$ has no solution.

Need $y^2-5y+3 < 0$ to change the sign of the LHS.

Need $(y - \frac{5}{2})^2 < 3.25$

$ -\sqrt{3.25} < y - \frac{5}{2} < \sqrt{3.25}$

$ -\sqrt{3.25} + \frac{5}{2} < y < \sqrt{3.25} + \frac{5}{2}$

$ 0.69722 < y < 4.3028$

so $y = [1,2,3,4]$