Is $(\log_2(\sqrt{5}+1))^3$ Greater Than $1+\log_2(\sqrt{5}+2)$?

[anemone]

Compare the numbers $X=(\log_2(\sqrt{5}+1))^3$ and $Y=1+\log_2(\sqrt{5}+2)$ and determine which is larger.

 

[Greg]

Compare the numbers $X=(\log_2(\sqrt{5}+1))^3$ and $Y=1+\log_2(\sqrt{5}+2)$ and determine which is larger.

Spoiler

The golden ratio (which I'll denote here as $\varphi$) satisfies $1.61<\varphi<1.62$ and is equivalent to$\dfrac{\sqrt5+1}{2}$, so we have

$$(\log_2(2\varphi))^3=(1+\log_2\varphi)^3=1+3\log_2\varphi+3\log_2^2\varphi+\log_2^3\varphi$$

Subtracting $1$ and approximating with $\sqrt2$ we have

$$X-1>\dfrac32+\dfrac34+\dfrac18=2.375$$

Also,

$$Y-1=\log_2(\sqrt5+2)=\log_2(2\varphi+1)<\log_24.24<\log_217-2<2.375$$

as required.

 

[anemone]

Thanks greg1313 for participating and thanks for your creative solution!(Cool)

I will wait for a bit before showing my solution and also the other insightful solution, just in case there are others who wanted to take part in this challenge.:)

 

[anemone]

My solution:

Spoiler

First, note that $5\gt 1$, which gives $2\sqrt{5}\gt 2$ and further translates into $5+2\sqrt{5}+1=(\sqrt{5}+1)^2\gt 8$, which implies $\sqrt{5}+1\gt 2^{\frac{3}{2}}$, taking base 2 logarithm of both sides of the inequality we get:

$\log_2(\sqrt{5}+1)\gt \dfrac{3}{2}$

$(\log_2(\sqrt{5}+1))^3\gt \left(\dfrac{3}{2}\right)^3=\dfrac{27}{8}$

On the other hand, we have

\(\displaystyle \frac{5}{2}\gt \sqrt{5}\)

\(\displaystyle \frac{9}{2}=\frac{5}{2}+2\gt \sqrt{5}+2\)

\(\displaystyle \therefore \log_2\left(\frac{9}{2}\right)\gt \log_2(\sqrt{5}+2)\)

\(\displaystyle \log_29 -1\gt \log_2(\sqrt{5}+2)\)

\(\displaystyle 1+ \log_29 -1\gt 1+\log_2(\sqrt{5}+2)\)

\(\displaystyle \log_29 \gt 1+\log_2(\sqrt{5}+2)\)

If we can prove \(\displaystyle \dfrac{27}{8}\gt \log_29\), then we can say $X\gt Y$.

Note that $2^{27}=(2^5)^{5+2} \gt 4(3^3)^5 \gt 3^{16}=9^8$, this suggest \(\displaystyle \dfrac{27}{8}\gt \log_29\) First, note that $5\gt 1$, which gives $2\sqrt{5}\gt 2$ and further translates into $5+2\sqrt{5}+1=(\sqrt{5}+1)^2\gt 8$, which implies $\sqrt{5}+1\gt 2^{\frac{3}{2}}$, taking base 2 logarithm of both sides of the inequality we get:

$\log_2(\sqrt{5}+1)\gt \dfrac{3}{2}$

$(\log_2(\sqrt{5}+1))^3\gt \left(\dfrac{3}{2}\right)^3=\dfrac{27}{8}$

On the other hand, we have

\(\displaystyle \frac{5}{2}\gt \sqrt{5}\)

\(\displaystyle \frac{9}{2}=\frac{5}{2}+2\gt \sqrt{5}+2\)

\(\displaystyle \therefore \log_2\left(\frac{9}{2}\right)\gt \log_2(\sqrt{5}+2)\)

\(\displaystyle \log_29 -1\gt \log_2(\sqrt{5}+2)\)

\(\displaystyle 1+ \log_29 -1\gt 1+\log_2(\sqrt{5}+2)\)

\(\displaystyle \log_29 \gt 1+\log_2(\sqrt{5}+2)\)

If we can prove \(\displaystyle \dfrac{27}{8}\gt \log_29\), then we can say $X\gt Y$.

Note that $2^{27}=(2^5)^{5+2} \gt 4(3^3)^5 \gt 3^{16}=9^8$, this suggest \(\displaystyle \dfrac{27}{8}\gt \log_29\) and we're hence done as we can conclude by now that $X\gt Y$.