Why is it trivial that $\left(1+\frac{2\ln3}{3}\right)^{-3/2}\leq\frac{2}{3}$?

Question

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.

It's trivial. The left hand side is 0.438556..., which is obviously smaller than 2/3. — DumpsterDoofus, Jan 28 '14 at 20:41

score 35 · Accepted Answer · answered Jan 28 '14 at 15:53

35

I imagine something along the lines of this, where each step is a gross simplification

$$ \left(1+\dfrac{2\ln3}{3}\right)^{-3/2} \leq \left(1+\dfrac{2}{3}\right)^{-3/2} = \left(\dfrac{5}{3}\right)^{-3/2} = \left(\dfrac{3}{5}\right)^{3/2} \leq \frac{3}{5} \leq \dfrac{2}{3}$$

answered Jan 28 '14 at 15:53

Your first step is incorrect. 2 ln 3 > 2. – Joel Rondeau Jan 28 '14 at 20:55
7

@JoelRondeau but x^(-3/2) is a decreasing function. – greggo Jan 28 '14 at 20:56
6

Thank you. Of course. Will leave that here so no one else makes the same mistake. – Joel Rondeau Jan 28 '14 at 20:57

score 17 · Answer 2 · answered Jan 28 '14 at 16:10

17

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.

answered Jan 28 '14 at 16:10

Christian Blatter

226,825

13

+1. "Trivial" does not mean "obvious" or "quickly proven". It means that, whatever amount of work is needed for the proof, it is uninteresting grunt work. Of course, being obvious makes a proof also trivial, but people forget that not everything trivial is also obvious. – rumtscho Jan 28 '14 at 18:43

score 12 · Answer 3 · edited Apr 13 '17 at 12:21

12

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.

edited Apr 13 '17 at 12:21

Community

1

answered Jan 28 '14 at 17:50

Will Orrick

18,220

score 3 · Answer 4 · answered Jan 29 '14 at 13:39

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 $\ln 3 > 1$

score 1 · Answer 5 · answered Jan 28 '14 at 16:07

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 $x>-1$).

Why is it trivial that $\left(1+\frac{2\ln3}{3}\right)^{-3/2}\leq\frac{2}{3}$?

5 Answers5