TITLE: If $(x^2-5x+4)(y^2+y+1)<2y$ for all real values of $y$ then what interval does $x$ belong to?
QUESTION [5 upvotes]: Okay so I'm stuck on this one for a long time :

If $(x^2-5x+4)(y^2+y+1)<2y$ for all real values of $y$ then what interval does $x$ belong to? 

My Try: Since $y$ can take any real value, I tried multiplying the first two terms and formed a quadratic equation in $y$ with its coefficients and the constant term in terms of $x$. Then I applied the two conditions as:

Coefficient of $y^2$ should be negative ; and 
Discrimnant should also be negative. 

Am I correct? What am I doing wrong? Is there a better approach to solve similar problems like this one? Please let me know.

REPLY [5 votes]: $$(x^2−5x+4)(y^2+y+1)<2y$$
$y^2+y+1$ is positive, because it is smallest at $y=-\frac{1}{2}$, and substituting that gives something positive.
Therefore we can divide by $y^2+y+1$ without changing the direction of the inequality.
$$x^2−5x+4<\frac{2y}{y^2+y+1}$$
We want this to be true for all values of $y$. In particular it has to be true when the$ \frac{2y}{y^2+y+1}$ is at its smallest, which is when it places the tightest restriction on the left hand side.
To minimise $\frac{2y}{y^2+y+1}$, you can look at where its derivative changes sign from negative to positive (which is at $y=-1$) or do the trick that DougM did in the other answer (setting it equal to $a$, and finding the smallest $a$ for which the quadratic in $y$ still has a solution). Either way, the minimal value is $-2$.
This gives:
$$x^2−5x+4<-2\\
(x-5/2)^2-25/4+4<-2\\
(x-5/2)^2<1/4\\
-1/2<x-5/2<1/2\\
2<x<3$$