# Triangle Inequality and bounds

I need help answering the following question:

Use triangle Inequality and reverse triangle inequality to find upper bound for $|(x^2-3)/(x-2)|$ if $x$ ranges over $|x-1| \lt \frac23$

I'm having trouble understanding the triangle inequality.

So far I have that $\frac13 \lt x \lt \frac53$

I just don't know how to apply that to find the bounds and my notes from class are unclear!

The main difficulty comes from the denominator. I would go that way, sinc we have an information about $\lvert x-1\rvert$: as $x^2-3=(x-1)^2+2(x-2)$, we can write $$\biggl\lvert\frac{x^2-3}{x-2}\biggr\rvert=\biggl\lvert\frac{(x-1)^2}{x-2}+2\biggr\rvert\le\biggl\lvert\frac{(x-1)^2}{x-2}\biggr\rvert+2.$$
by the triangle inequality. Now $(x-1)^2\le \dfrac49$ and, by the reverse triangle inequality: $$\lvert x-2\rvert=\lvert(x-1)-1\rvert\ge\biggl\lvert\frac23-1\biggr\rvert=\frac13$$ whence $\;\biggl\lvert\dfrac{x^2-3}{x-2}\biggr\rvert\le\dfrac{\dfrac49}{\dfrac13}+2=\dfrac{10}3$.