<s>[INST] Finding the maximum value of $y=\frac{x^2+x-1}{x^2-x+1}$ without calculus The problem is:

Find the maximum value of
$$y=\frac{x^2+x-1}{x^2-x+1}$$ assuming $x\geq 0$. I am only interested in pre-calculus solutions.

Using calculus it is easy to see that the maximum is $y=\frac{5}{3}$ occuring at $x=2$.
Other forms are
$$y=\frac{(x+\frac{1}{2})^2-\frac{3}{4}}{(x-\frac{1}{2})^2+\frac{5}{4}}$$
$$y=\frac{(\sqrt{x}-\frac{1}{\sqrt{x}})(\sqrt{x}+\frac{1}{\sqrt{x}})+1}{(\sqrt{x}-\frac{1}{\sqrt{x}})^2+1}$$
 [/INST] A simple approach is to rewrite it as a quadratic for $x$
$$ (y-1)x^2 - (y+1)x + (y+1)=0$$
And require that the discriminant be not negative
$$(y+1)^2 - 4(y+1)(y-1) \ge 0$$
$$(y - \frac{5}{3})(y+1) \le 0$$
Which gives us the range for $y$.  To be thorough, it's also good to check that the maximum value for $y$ does not correspond to a zero in the denominator of the original fraction.  In this case, it doesn't.
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