2016 'N 2018

Bernoulli's inequality is stated as follows: $1+nx \leq (1+x)^n$ .

Now, Bernoulli's approximation is just like that, only replacing the $\leq$ sign with a $\approx$ sign, i.e. $1+nx \approx (1+x)^n$ .

Now, this approximation works when $n \approx 1$ , which is exactly what is happening in the problem, when we rewrite it like this

$\left\lfloor\dfrac{\sqrt[2018]{2016^{2016}} + \sqrt[2016]{2018^{2018}}}{2}\right\rfloor = \left\lfloor\dfrac{(1+2015)^\frac{2016}{2018} + (1+2017)^\frac{2018}{2016}}{2}\right\rfloor$

Now, $(1+2015)^\frac{2016}{2018} + (1+2017)^\frac{2018}{2016} \approx 1+\frac{(2015)(2016)}{2018} + 1 + \frac{(2017)(2018)}{2016} = 4034.003$ . Thus,

$\left\lfloor\dfrac{\sqrt[2018]{2016^{2016}} + \sqrt[2016]{2018^{2018}}}{2}\right\rfloor \approx \left\lfloor\dfrac{4034.003}{2}\right\rfloor = \boxed{2017}$

Which is our final answer since we have $(1+2015)^\frac{2016}{2018} + (1+2017)^\frac{2018}{2016} \geq 1+\frac{(2015)(2016)}{2018} + 1 + \frac{(2017)(2018)}{2016}$

by Bernoulli's inequality.

I solve it because i think 2016/2018 and 2018/2016 are near 1 and I just add 2016+2018 then divide it by 2 so luckily it's right