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?

Solution 1:

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.

Solution 2:

$$1+{1\over2018}<1+{1\over2000}=1.0005$$ $$\left(1+{1\over2018}\right)^{2018}<(1.0005)^{2048}$$

The right hand side can be evaluated by repeated squaring, and, since there is plenty of leeway, you can simplify the calculations by rounding up as you go. Since $(1.0005)^{2048}<3,$ this calculation should be well within you students' abilities.

Solution 3:

Note that $2019^{2048}<2018^{2049}$ implies that $$2019^{2018}2019^{30}=2019^{2048}<2018^{2049}<2018^{2019}2019^{30},$$ which implies that $2019^{2018}<2018^{2019}$. We look at $2048$ in the exponent because it is a power of $2$.

Claim: $2019^{2048}<2018^{2049}$.

Proof: $$2019^{2048}-2018^{2048}=$$ $$(2019^{1024}+2018^{1024})(2019^{512}+2018^{512})...(2019^{2}+2018^{2})(2019+2018)(2019-2018)$$ by repeatedly factoring differences of squares. Each term of the form $2019^i+2018^i<2 \cdot 2019^i$, so taking each of these inequalities into account, we get that $$2019^{2048}-2018^{2048}<2^{10}\cdot 2019^{2047}$$ since $1+2+4+8+...+512+1024=2^{11}-1$. Then we combine terms with like bases to get that $$2019^{2048}-1024\cdot 2019^{2047}=995\cdot 2019^{2047}<2018^{2048}.$$ We now multiply both sides by $2018$ to get $$2019^{2048}<995\cdot 2018\cdot 2019^{2047}<2018^{2049},$$ which is the desired result.

Solution 4:

Borrowing from a recent answer of mine (it proves more than is needed here, and it avoids using the Binomial Theorem, but it does use induction, and it is arguably a bit too complicated, even for its original purpose, so I've simplified it considerably for the present application):

Define $$a_n = \left(1+\frac{1}{n}\right)^n \quad (n \geqslant 1). $$

If $x \geqslant y > 0$, and $n$ is a positive integer, then $$ x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + y^{n-1}) \geqslant n(x - y)y^{n-1}. $$ This can be convincingly justified without an explicit use of induction, just a plausible use of ellipsis.

Therefore, for $n > 1$, \begin{align*} a_n - a_{n-1} & = \left(1+\frac{1}{n}\right)^n \! - \left(1+\frac{1}{n-1}\right)^{n-1} \\ & = \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \left[ \left(1+\frac{1}{n-1}\right)^{n-1} \!\! - \left(1+\frac{1}{n}\right)^{n-1}\right] \\ & \leqslant \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{1}{n^2}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{a_n}{(n+1)^2}, \end{align*} whence $$ a_n \leqslant a_{n-1}\left(1 - \frac{1}{(n+1)^2}\right)^{-1} \quad (n > 1). $$ But, for $n > 1$, $$ \left(1 - \frac{1}{(n+1)^2}\right)^{-1} \!\! = 1 + \frac{1}{(n+1)^2 - 1} < 1 + \frac{1}{n-1} = \frac{n}{n-1}, $$ therefore $$ \frac{a_n}{n} < \frac{a_{n-1}}{n-1} \quad (n > 1), $$ and - again without explicit use of induction - one can deduce: $$ \frac{a_n}{n} \leqslant \frac{a_3}{3} = \frac{64}{81} < 1 \quad (n \geqslant 3), $$ whence: $$ (n + 1)^n < n^{n+1} \quad (n \geqslant 3). $$