# How to prove this inequality with Big O term?

Let $$s= s_0-\zeta_0^{-1/2}b^{-1/6}(1+\mathcal{O}(\sqrt{b}))$$ where $$\zeta_0 = \left(\frac{3\pi}{4}\right)^{2/3}$$ and $$s_0 = b^{-2/3}\zeta_0.$$ Note that here $$b$$ is a parameter.

We define $$\omega(s) = \exp\left(\frac{2}{3}s_{+}^2\right)$$ where $$s_{+} = \max(0,s).$$ I want to show that for some constant $$C$$ we have that, $$\omega(s) \leq C\exp\left(\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\right)\leq C\exp\left(-\frac{1}{2\sqrt{b}}\right)\Sigma_{b}^{-1},$$ where $$\Sigma_{b}^{-1}=\exp\left(\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\right).$$

For the second inequality does not make sense to me as $$\exp\left(-\frac{1}{2\sqrt{b}}\right)\leq 1$$, but I might be wrong.

For the first inequality, $$\frac{2}{3}s_{+}^{2}\leq \frac{2}{3}|s|^2\leq \frac{4}{3}|s_0|^2+\frac{8}{3}(\zeta_0^{-1}b^{-1/3} + C\zeta_0^{-1}b^{2/3})$$ $$=\frac{4}{3}\left(\frac{3\pi}{4b}\right)^{4/3}+\frac{8}{3}\left(\left(\frac{4}{3\pi\sqrt{b}}\right)^{2/3} + C\left(\frac{4b}{3\pi}\right)^{2/3}\right)$$

where I used the inequality $$|a+b+c|^2\leq 2|a|^2+4(|b|^2+|c|^2).$$ I cannot see how I reduce this expression to derive the first inequality. Any hints would be much appreciated.

The first inequality is equivalent to $$\frac 23s_+^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$$ for some constant $$D$$. Since $$\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\ge -\frac 1{4\pi}$$ for each $$b\ge 0$$, if $$D\ge -\frac 1{4\pi}$$ then the inequality holds for $$s_+=0$$. So it remains to consider an inequality
$$\frac 23s^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$$
$$\frac 23\left(b^{-2/3}\zeta_0-\zeta_0^{-1/2}b^{-1/6}(1+\mathcal{O}(\sqrt{b}))\right)^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$$
For big $$b$$ this inequality can fail. For instance, when $$\mathcal{O}(\sqrt{b})$$ term equals $$-1-\sqrt{b}$$, then when $$b$$ tends to infinity, the left hand side of the inequality tends to infinity, whereas the right hand side is bounded. For small $$b$$ the inequality fails too. Indeed, when $$b$$ tends to zero, the left hand side of the inequality grows as $$b^{-4/3}$$, whereas the right hand grows as $$b^{-1}$$.