Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. This inequality is equivalent to
$$
2018^{1/2018}>2019^{1/2019}
$$
One of my 'High school' student asked why the inequality is true. The whole class became interested in the problem.The demonstration that such inequality is true, using calculus, can be found here. But my students are not familiar with calculus.
I can also show by induction and Newton's binomial formula that $ n^{(n + 1)}> (n + 1)^n $, to $ n> 3$, but my students are not familiar with mathematical induction.
Another limitation of my students is that they have not yet learned the Newton's binomial formula.
How to prove the inequality $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without calculus? That is how to prove this inequality for High school students without using Newton's binomial formula?
Best Answer
I would try to motivate this by showing that $f(x) = x^{1/x}$ is a monotone function for reasonably small $x$.
Another approach is to note that $$ \frac{2019^{2018}}{2018^{2018}} = \left(1 + \frac{1}{2018}\right)^{2018} $$ and so your inequality is equivalent to showing $$ \left(1 + \frac{1}{2018}\right)^{2018} < 2018, $$ which does not sound very far-fetched, since LHS is close to $e$...
UPDATE
Please see saulspatz's answer for how to prove this last claim with a hand computation only.