<s>[INST] Use quadratic formula to find upper and lower limits of an expression 
Using quadratic formula show that $\frac{x^2-x+1}{x^2+x+1}$ lies
  between $3$ and $\frac{1}{3}$ for all real values of $x$.

Let $\frac{x^2-x+1}{x^2+x+1}=y$, then $\frac{x^2+1}{-x}=\frac{y+1}{y-1}$
$\therefore (y-1)x^2+(y+1)x+(y-1)=0$, if $x$ is real then 
$(y+1)^2-4(y-1)(y-1)\geq 0$
$-3y^2+10y-3 \geq 0$
$(y-3)(-3y+1) \geq 0 \Rightarrow y>3 \text{ AND } y <\frac{1}{3}$
What am I doing wrong?
 [/INST] You said:

$$(y-3)(-3y+1) \geq 0 \Rightarrow y>3 \text{ AND } y <\frac{1}{3}$$

No, this isn't true. We have three options:


*

*Both factors are positive $y>3 \text{ AND } y <\frac{1}{3}$, no $y$ statistifies this. 

*Both factors are zero $y=3 \text{ AND } y=\frac{1}{3}$, no $y$ statistifies this. 

*Both factors are negative $y<3 \text{ AND } y >\frac{1}{3}$, so $\frac{1}{3}<y<3$.


In general, this isn't how you should solve these kind of equations. Instead, solve the equation (not the inequality, so change $>, \geq, < , \leq$ to $=$) and look to the graph.
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