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.
 A: I imagine something along the lines of this, where each step is a gross simplification
$$ \left(1+\dfrac{2\ln3}{3}\right)^{-3/2}
\leq  \left(1+\dfrac{2}{3}\right)^{-3/2}
=  \left(\dfrac{5}{3}\right)^{-3/2}
= \left(\dfrac{3}{5}\right)^{3/2}
\leq \frac{3}{5}
\leq
\dfrac{2}{3}$$
A: We can see everything is positive, so just take reciprocal and square both sides:
$\left(1+ \frac{2 \ln 3}{3}\right) ^ {\frac{3}{2}} \ge \frac{3}{2}$
$\left(1+ \frac{2 \ln 3}{3}\right) ^ 3 \ge \frac{9}{4}$
Bring down the three inside the bracket (first step of expansion), since everything is positive the extra terms will be positive:
$1 + 2 \ln 3 + \text{stuff} \ge 1+ \frac{5}{4}$
Which is clearly true since $\ln 3 > 1$
A: It's not trivial if you have to prove this "with bare hands"; see Hurkyl's answer. 
But it's trivial if you have a pocket calculator at your disposal. So I'm interpreting the phrase as follows: The author was on his way to heavier and more important things, and he didn't want to interrupt his argument with a proof of this little fact. Depending on circumstances and the envisaged audience he could have made a "Lemma" out of it.
A: Taking reciprocals of both sides gives the equivalent statement
$$
\left(1+\frac{2\ln3}{3}\right)^{3/2}\ge\frac{3}{2},
$$
which is fairly obvious once you realize that $\ln3>1.$  The latter follows from the fact that $e<3,$ which is not hard to see.
A: 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 $x>-1$).
