Solving inequality by investigating non-negativity of terms

Question

I've stumbled upon this inequality in a larger proof

$$\ln \left( \frac{ x^2 a / b - b/2}{x} \right) + \left(\frac{ x^2 a}{ b} - \frac{b}{2} \right)^2 > \ln \left( \sqrt{\frac{2}{\pi}} \frac{1}{d}\right)$$

where $a, b, d \in \mathbb{R}$ are some positive constants and we're solving for $x$. We are also given $a \leq 1 \leq b$ (and $d$ is very small, which might be important?).

It seems that there is no closed-form solution (I've tried a couple and failed), so any good approximation should do.

The authors say

Let us write $x = c \cdot b /a$, we wish to bound $c$. We begin by finding the conditions under which the first term is non-negative.

Can someone enlighten me

1. why introducing $c$ is useful, and
2. why non-negativity of the first term is helpful in solving this inequality in the first place?

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Edit: We can even simplify the RHS, as this is a constant anyway:

$$\ln \left( \frac{ x^2 a / b - b/2}{x} \right) + \left(\frac{ x^2 a}{ b} - \frac{b}{2} \right)^2 > D$$

(and $D$ would be a positive number, maybe large, but arbitrary)

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Edit 2: The RHS constant is actually not that large, typical $D$ values would be somewhere between $10.0$ and $60.0$ (if that helps the proof?)

Answer 1

Your inner questions:

• 1. if you introduce $x=c\,b/a$ the logarithm of the LHS becomes $\log\left(c-\frac{a}{2c}\right)$ which is easier to analize since, for having a well define logarithm, its argument must be positive, so you can rise the restriction $c-\frac{a}{2c}\geq 0$ that it could be easier to work than the original.
• 2. by the same argument of having logarithms with positive arguments, you could made the LHS to look as $$ \log\left(\frac{x^2\,a}{b}-\frac{b}{2}\right)-\log(x)+\left(\frac{x^2\,a}{b}-\frac{b}{2}\right)^2$$ so for having well defined logarithm you will need that $x\geq 0$ and $$\frac{x^2\,a}{b}-\frac{b}{2}\geq 0 \Rightarrow x^2 \geq \frac{b^2}{2a}$$. Now, since the second term of the original inequality is always non-negative because is a "squared number", the non-negativity of the arguments is required for having well defined logarithms (on converse, the inequality is not solvable). Note I have arbitrarily assume that $a>0$ and $b>0$ for this example, otherwise, the inequality could change direction... you must keep track of this issues and split the problem on each possible scenario. My point here is only to note what arguments can been built to figure out the final answer (the same stands for the following explanation).

Now for the overall inequality, it could be transform into: $$ \log\left(\frac{x^2\,a}{b}-\frac{b}{2}\right)-\log(x)+\log\left(e^{\left(\frac{x^2\,a}{b}-\frac{b}{2}\right)^2}\right)-\log\left(\sqrt{\frac{2}{\pi}}\frac{1}{d}\right)>0$$ $$\Rightarrow \left(\frac{x^2\,a}{b}-\frac{b}{2}\right)\,e^{\left(\frac{x^2\,a}{b}-\frac{b}{2}\right)^2}\,\frac{d}{x} \sqrt{\frac{\pi}{2}}>1$$ since to have a positive logarithm, its argument must be higher than one... but here you can see than the variable $x$ is over and multiplying an exponential, so a close-form it can't be found.

Now, since already $x\geq 0$: $$ \left(\frac{x^2\,a}{b}-\frac{b}{2}\right)\,e^{\left(\frac{x^2\,a}{b}-\frac{b}{2}\right)^2}\,d\sqrt{\frac{\pi}{2}}>x$$ And since $u=\left(\frac{x^2\,a}{b}-\frac{b}{2}\right)\geq 0$ I have that $$ ue^{u^2}d\sqrt{\frac{\pi}{2}}>\sqrt{\frac{b}{a}\left(u+\frac{b}{2}\right)}>0$$ which, after more manipulations ($z=u^2 \geq 0$): $$ze^z > \frac{\sqrt{z}}{d}\sqrt{\frac{2}{\pi}}\sqrt{\frac{b}{a}\left(\sqrt{z}+\frac{b}{2}\right)}$$ which could be numerically investigated through the Lambert W function and noting later that the RHS is: $$\frac{\sqrt{z}}{d}\sqrt{\frac{2}{\pi}}\sqrt{\frac{b}{a}\left(\sqrt{z}+\frac{b}{2}\right)} = \frac{x}{d}\sqrt{\frac{2}{\pi}}\sqrt{u}$$