Let be $I = [a,b]$ and $c$ the center of this interval. I need to show that for $x \not \in 2I$
$$|\log\left(\frac{|x-a| |x-b|}{|x-c|^ 2}\right)|<4 \frac{|b-a|^2}{|x-c|^2}$$
Where $x \not \in 2I$ means that $x \not \in [c-(b-a),c + (b-a)]$ My try:
$\log\left(\frac{|x-a| |x-b|}{|x-c|^ 2}\right) \leq \log\left(\frac{|x-a| |x-b|}{|x-c|^ 2} + 1\right)$ then by applying Taylor for $|t|=\frac{|x-a| |x-b|}{|x-c|^ 2} \leq \frac{1}{2}$
$$\left|\log\left(\frac{|x-a| |x-b|}{|x-c|^ 2}\right) \right| \leq |t + R_1(t)| $$ where $|R_1(t)|\leq2 |t|^ 2$. That implies:
$$\left|\log\left(\frac{|x-a| |x-b|}{|x-c|^ 2}\right)\right| \leq \frac{|x-a| |x-b|}{|x-c|^ 2} + 2\frac{|x-a| |x-b|}{|x-c|^ 2} \cdot \frac{|x-a| |x-b|}{|x-c|^ 2}$$
Using the hypothesis that $x \not \in 2I$ (WHY?):
$$|x-a|< |b-a| \text{ and } |x-b|< |b-a| \text{ (*) }$$
$$\leq 2 \frac{|b-a|^2}{|x-c|^2} + 4 \frac{|b-a|^2}{|x-c|^2} \cdot \frac{1}{2} =4 \frac{|b-a|^2}{|x-c|^2}$$
I'm not sure about this proof because I don't know if it's correct to assume (*) given that $x \not \in 2I$. Could someone give me a hint? I need to use Taylor for this proof