Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. 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?


Notice that $2018^x > x+1$ for any $x>0$ (plot each function if need be to convince your high school students).
Then $x = \frac{2019}{2018} -1 > 0$ gives 
$$\frac{2018^{\frac{2019}{2018}}}{2018} > \frac{2019}{2018}$$
which gives the desired result after clearing denominators and taking each side to the $2018$ power.
Note: This argument is more commonly used on comparing $e^2$ to $2^e$.
