I have the difference between the following log expressions and I am trying to bound the difference,
$$F= \log \left(1+ \left(2+\frac{1}{\sqrt{2}}\right)^2 x^2\right) - \log \left(1+ \left(1-\frac{1}{\sqrt{2}}\right)^2 x^2\right) $$
Can I say that $$F \leq \log\left(1+ \frac{ \left(2+\frac{1}{\sqrt{2}}\right)^2}{\left(1-\frac{1}{\sqrt{2}}\right)^2}\right)$$
Is this the tightest bound that exists?
Thanks in advance.
On
$f(a) =\log (1+ax^2) - \ln a\Rightarrow f'(a) = \dfrac{x^2}{1+ax^2}- \dfrac{1}{a}< 0\Rightarrow \text{ since } b=\left(2+\dfrac{1}{\sqrt{2}}\right)^2 > \left(1-\dfrac{1}{\sqrt{2}}\right)^2=a\Rightarrow f(a) < f(b)\text{ and} \ln\left(\dfrac{b}{a}\right) \leq \dfrac{b}{a}-1 \Rightarrow \text{the inequality holds and the tighter bound is} \dfrac{b}{a}-1$
Set $\alpha=(2+\frac{1}{\sqrt{2}})^2$ and $\beta=(1-\frac{1}{\sqrt{2}})^2$, each positive, with $\alpha>\beta$.
We have $$F= \log\left(\frac{1+\alpha x^2}{1+\beta x^2}\right)$$ For all $x$, the fraction is in $[1,\frac{\alpha}{\beta})$. Further, as $x\to\infty$, the fraction gets closer and closer to the right endpoint. Hence the tightest possible bound for $F$ that will be valid for all $x$ is $$F\le \log \frac{\alpha}{\beta}=\log \alpha -\log \beta$$
If you don't like logs we can use the Taylor approximation $$\log x=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+\cdots$$
We set $x=\frac{\alpha}{\beta}$. The series is alternating, so if we truncate after a negative term we get a succession of upper bounds: $$F\le \frac{\alpha}{\beta}-1$$ $$F\le \frac{\alpha}{\beta}-1-\frac{(\frac{\alpha}{\beta}-1)^2}{2}+\frac{(\frac{\alpha}{\beta}-1)^3}{3}$$ etc. Note that the first bound is already sharper than the one in the OP.