The power of triples #2

Let's start with the immediately obvious inequality: $2^5\cdot111>3^4$ . (Note that the left-hand-side is clearly more than $100$ , while the right-hand-side is only $81$ .)

Multiplying by $111^4$ , we get $2^5\cdot111^5>3^4\cdot111^4$ , or $222^5>333^4$ . Raising to the $111\text{st}$ power, we get: $222^{555}>333^{444}$ We prove that $333^{444}>444^{333}$ in a similar way; this time, start off by noticing that $3^4\cdot111>4^3$ , and follow the same argument. Again, $444^{333}>555^{222}$ can be proven using the same argument if you start with $4^3\cdot111>5^2$ .

Since the second to fourth numbers are all less than $666^{444}$ , it suffices to show that $666^{444}<222^{555}$ , or , after taking 111th roots, show that $666^4<222^5$ . But $666^4=3^4222^4<222^5$ since $3^4=81<222.$

My first thought, when I saw the question, was to take the logarithm of everything. So... $log_a(222^{555}) = 555\:log_a(222) = 555(log_a(111)+log_a(2))\\ log_a(333^{444}) = 444\:log_a(333) = 444(log_a(111)+log_a(3))\\ log_a(444^{333}) = 333\:log_a(444) = 333(log_a(111)+log_a(4))\\ log_a(555^{222}) = 222\:log_a(555) = 222(log_a(111)+log_a(5))$ It's entirely up to me to which base to take logs, so let $a=2$ .

First, we want to compare $555(log_2(111)+log_2(2))$ and $444(log_2(111)+log_2(3))$ . This is equivalent to comparing $log_2(111) + 5\:log_2(2)$ and $4\:log_2(3)$ . Now, $6<log_2(111)<7$ , making the former equal to at least $11$ . Moreover, $log_2(3)<2$ , so the latter is at most $8$ , whence $222^{555} > 333^{444}$ .

The same argument holds for $222^{555}$ and $444^{333}$ ; now you have $2\:log_2(111)$ which dominates the inequality, and the former is larger again.

Similarly for $555^{222}$ .

$\Large 222^{\frac{555}{444}}=856>555 ~~So~~222^{555} ~ largest.\\ \text {Root of the biggest power and still bigger than the biggest number-base.}$

Another cute way which generalizes the problem: prove that $(111a)^{7-a}>(111(a+1))^{6-a}$ . It's equivalent to $111(a+1)>(1+\frac{1}{a})^{7-a}$ . Note that $1+\frac{1}{a}<2$ , so if the opposite of our inequality was true, then $111(a+1)<2^{7-a}$ which is clearly false for $a \ge 1$ . Thus the function $(111a)^{7-a}$ is strictly decreasing for $a \ge 1$ .