<s>[INST] Why is it trivial that $\left(1+\frac{2\ln3}{3}\right)^{-3/2}\leq\frac{2}{3}$? Can someone tell me why $$\left(1+\dfrac{2\ln3}{3}\right)^{-3/2}\leq\dfrac{2}{3}$$ is trivial because for me its not and I will need to do the calculation to see it.
 [/INST] You have that
\begin{align*}
\left(1+\frac{2\ln3}{3}\right)^{-\frac{3}{2}} &= e^{ -\frac{3}{2}\ln\left(1+\frac{2}{3}\ln3\right) } \\
&=e^{ \frac{3}{2}\ln\frac{1}{1+\frac{2}{3}\ln3} } \\
&\leq e^{ \frac{3}{2}\ln\left( 1-\frac{1}{2}\cdot\frac{2}{3}\ln3 \right) } \\
&\leq e^{ -\frac{3}{2}\frac{1}{3}\ln 3 } = e^{ -\frac{1}{2}\ln 3 } = \frac{1}{\sqrt{3}} \\
&\leq \frac{2}{3}
\end{align*}
where was used the "fact" that $\frac{1}{1+\frac{2}{3}\ln3}\leq 1-\frac{1}{2}\cdot\frac{2}{3}\ln3$ (proving it is not hard) and $\ln(1+x)\leq x$ (for $x>-1$).
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