Prove: $\log_2(x)+\log_x(y)+\log_y(8)\geq \sqrt[3]{81}$

Question

Prove that for every $x$,$y$ greater than $1$: $$\log_2(x)+\log_x(y)+\log_y(8)\geq \sqrt[3]{81}$$

What I've tried has got me to:

$$\frac{\log_y(x)}{\log_y(2)}+\log_x(y)+3\log_y(2)\geq \sqrt[3]{81}$$

I didn't really get far.. I can't see where I can go from here, especially not what to do with $ \sqrt[3]{81}$.

This is taken out of the maths entry tests for TAU, so this shouldn't be too hard.

Answer 1

We have $$\log_2x+\log_xy+\log_y8=\frac{\log x}{\log2}+\frac{\log y}{\log x}+\frac{3\log2}{\log y}$$

and since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.

Now use AM-GM:

$$\sqrt[3]{\frac{\log x}{\log2}\cdot\frac{\log y}{\log x}\cdot\frac{3\log2}{\log y}}\le\frac{\frac{\log x}{\log2}+\frac{\log y}{\log x}+\frac{3\log2}{\log y}}3$$ giving $$\frac{\log x}{\log2}+\frac{\log y}{\log x}+\frac{3\log2}{\log y}\ge3\sqrt[3]3=\sqrt[3]{81}$$ as required.

Answer 2

Hint: Show that $a+\frac{b}{a}+\frac{3}{b}\geq \sqrt[3]{81}$ for all $a,b>0$, using AM-GM.

Answer 3

Let's convert everything to $\log$ base $2$ so we have a common something to work with:

$$\log_2 x + \dfrac {\log_2 y}{\log_2 x} + \dfrac {\log_2 8}{\log_2 y} \ge \sqrt [3]{81} = 3 \sqrt[3]{3}$$

Now let $a = \log_2 x$ and $b = \log_2 y$. The condition $x, y > 1$ implies that $a, b > 0$. So now we have to prove

$$a + \dfrac ba + \dfrac {3}{b} \ge3 \sqrt[3]{3}$$

The cube root in there especially may remind us of AM-GM with $3$ terms. And indeed, the inequality $\text{AM}\left(a, \dfrac ba, \dfrac 3b\right) \ge \text{GM}\left(a, \dfrac ba, \dfrac 3b\right)$ gives exactly what is desired.

Answer 4

Two things:

$\log_a b = \frac 1{\log_b a}$ and $\frac {\log_b c}{\log_b a} = \log_a c$ so $\log_a b\log_b c = \frac {\log_b c}{\log_b a} = \log_a c$.

And AM-GM says $\frac {a + b+ c}3 \ge \sqrt[3]{abc}$.

So.....

$\frac {\log_2 x + \log_x y + \log_y 8}3 \ge \sqrt[3]{\log_2 x \log_x y \log_y 8} = \sqrt[3]{\log_2 8}$