The power of triples #3

The following proof is more analytic and reduces the number of times you have to use the calculator. Let us introduce the function $f(x)=x^{1110-x}$ where $0<x<1110$ . What we have to find is the largest of the values ${f(123), f(234), ..., f(789)}$ . We claim that the largest of these values is $f(234)= 234^{876}$ . The proof of our claim can be reduced to verify that $f(x)$ is decreasing for $x\geq 234$ and $f(123)<f(234).$ The inequality is very easy to check by comparing the natural logarithms of the two numbers. To prove that $f(x)$ is decreasing on the given interval is equivalent to prove that $g(x)=\ln f(x) =(1110-x) \ln x$ is decreasing on the interval. This can be obtained by showing that $g'(x)= \frac{1110}{x} -1 -\ln(x)$ is negative for $x\geq 234$ and this follows from the fact that $g''(x)= -\frac{x+1110}{x} <0$ for any $x\geq 234$ and $g'(234)\approx -1.71 <0.$

$\Large \color{#D61F06}{ (a)~~ 123^{\frac{987}{876} }=226<234~~~~~~~\implies~(2)> (1)\\\Large (b)~~ 234^{\frac{876}{765}}=520>345~~~~~~~\implies~(2)> (3)}\\ \Large (c)~~ 234^{\frac{876}{654}}=1490>456~~~~\implies~(2)> (4)\\\Large \color{#D61F06}{ (d) ~~234^{\frac{876}{543}}=6639>567~~~~\implies~(2)> (5)}\\\Large (e) ~~234^{\frac{876}{765}}=63715>678~~\implies~(2)> (6)\\\Large (f)~~234^{\frac{876}{765}}=2920946>789~~~\implies~(2)> (7)\\ \huge \color{#3D99F6}{ 234^{876}}\\ ~~~~~~~\large e.g.~234^{\frac{876}{654}}=1490~is~termed~as~a~ ratio.\\\large \text{For ratios, only integers are used since this is sufficient.}\\~~~\\\\ \color{#20A900}{\text{Short cut:- Compare (3) and (4) with step (b) above. Exponent of (4) }\\\text{is less than that of (3) and in (c) the rartio we get is 520>456 the base}\\ \text{of (4), so (4)<(2). Similarly, in (d) ratio 6639> bases of (6) and (7) }\\ \text{and their exponents are less than 543 used in (d).Hence (6) and (7)}\\ \text{are less than (2). SO only steps (a),(b) and (d) are sufficient.} }$

... or use logarithms.