Proving that $3^{(3^4)}>4^{(4^3)}$ without a calculator

Is there a slick elementary way of proving that $3^{(3^4)}>4^{(4^3)}$ without using a calculator?

Here is what I was thinking: $$4^4=256>243=3^5,$$ hence $$4^{4^3}=4^{64}=(4^4)^{16}=(3^5)^{16}\cdot\left(\dfrac{256}{243}\right)^{16}=3^{80}\cdot\left(\dfrac{256}{243}\right)^{16}<3^{81}=3^{3^4}\,.$$ It is possible to prove that $\left(\dfrac{256}{243}\right)^{16}<3$ without a calculator by making a comparison such as $\dfrac{256}{243}=1+\dfrac{13}{243}<1+\dfrac{15}{240}=\dfrac{17}{16}$ and $\left(1+\dfrac{1}{16}\right)^{16}<e$ so $\left(\dfrac{256}{243}\right)^{16}<\left(1+\dfrac{1}{16}\right)^{16}<e<3$.

Is there a more elegant elementary proof? Ideally I'm looking for a proof that doesn't rely on calculus.

Source of problem: I made up this question but it is inspired by a similar question that appeared in the British Mathematical Olympiad round 1 in 2014.

Solution 1:

$\Large {3^{3^4} \over 4^{4^3}} = {3^{81} \over 4^{64}} = {3 \over \left({256\over243}\right)^{16}} > {3 \over (1+{13\over243})^{243\over13}} > {3 \over e} > 1$

Solution 2:

Your problem is equivalent to proving that $\log_2{3} > \frac{128}{81}$.

One approach is to calculate that $3^{12} = 81^3 = 531441 > 2^{19} = 2^{10} \times 2^9 = 1024 \times 512 = 524288$. Admittedly, this is a bit tedious with pencil-and-paper arithmetic, but it's doable. From this, we get $\log_2{3} > \frac{19}{12}$.

Separately, calculate that $\frac{19}{12} = \frac{513}{324} > \frac{128}{81} = \frac{512}{324}$.

Combining these results, using the fact that the > operator is transitive, we get $\log_2{3} > \frac{128}{81}$, Q.E.D.

Solution 3:

$3^4-4^3=17$ therefore we can rewrite the inequality as:

$$3^{17} > \left(\frac{4}{3}\right)^{4^3}$$

we have $\left(\frac{4}{3}\right)^8 < 10$, hence it suffices to show:

$$3^{17} > 10^8$$

or

$$3 \cdot 81^4 > 10^8$$

which is satisfied if:

$$3 \cdot 80^4 > 10^8$$

or

$$3 \cdot 2^{12} \cdot 10^4 > 2^{4} \cdot 5^4 \cdot 10^4$$

$$3 \cdot 2^8 > 5^4$$

i.e.

$$768 > 625$$

ADDENDUM

$4^8=4^4 \cdot 4^4 = 256 \cdot 256$ and $3^8=3^4 \cdot 3^4 = 81 \cdot 81$, it is not too difficult to make the two multiplications and then the division.

Also, if you are a programmer, you know already that $4^8 = 2^{16} = 65536$, so you just need to make $81 \times 81$ and then the division. Rather, you don't need any division because you see immediately $81 \cdot 81 = 6561 \gt 65536 / 10$.

Anyway, I admit, I have used the calculator :-)