The question specifically asks for an

*algebraic* method, so I'm not sure why the thread has been moved to the Calculus forum.

**Mr F says: Post #3.**
Having said that, it doesn't seem to be that easy to show this result by purely algebraic means. Suppose we start by looking at a simpler result:

*The function \(\displaystyle y = x + \frac4x\) cannot take values between –4 and +4.* To see that, notice that \(\displaystyle y^2 = \left(x+\tfrac4x\right)^2 = x^2 + 8 + 16x^{-2} = \left(x-\tfrac4x\right)^2 + 16.\) Thus \(\displaystyle y^2\geqslant 16\), which means that y cannot lie between –4 and +4.

Now let's try to use a similar method on the function \(\displaystyle y=\frac{2x(x+3)}{x-1}\). At this point, I will cheat slightly, by using the helpful graph linked to in Mr F's first comment above. That makes it look as though y cannot take values between 2 and 18, or in other words between 10–8 and 10+8. So we should try to prove that \(\displaystyle |y-10|\geqslant8.\)

The first step is to calculate

\(\displaystyle \begin{aligned}y-10 = \frac{2x(x+3) - 10(x-1)}{x-1} &= \frac{2(x^2-2x+5)}{x-1} \\ &= \frac{2\bigl((x-1)^2+4\bigr)}{x-1} = 2{\color{red}\bigl(}(x-1) + 4(x-1)^{-1}{\color{red}\bigr)}.\end{aligned}\)

Apart from having \(\displaystyle x-1\) in place of x, the expression in the red parentheses looks just like \(\displaystyle x+\tfrac4x.\) So you should be able to complete the argument from there.