TITLE: How to show :$\int_0^1 \left[\left(1-x^{2018}\right)^{1\over 2020}- \left(1-x^{2020}\right)^{1\over 2018} \right] dx \lt \frac {2018}{2020}$
QUESTION [7 upvotes]: I tried to solve this problem as $$0\lt x \lt 1 \implies 0\lt x^{2018} \lt 1 \implies 0\lt (1-x^{2018}) \lt 1$$
this means
$$\int_0^1 \left[\left(1-x^{2018}\right)^{1\over 2020}- \left(1-x^{2020}\right)^{1\over 2018} \right] dx \lt   \int_0^1 \left[\left(1-x^{2018}\right)^{1\over 2020}\right] dx \lt \int_0^1 1^{{1\over 2020}} dx$$ 
as inequality can be integrated.
So I have come as far as proving the given  expression is less than 1 but I cannot proceed further.
Can someone show me how to proceed?
(Please try give answers which can be understood by those in elementary calculus courses)

REPLY [5 votes]: You are thinking too hard.
Consider the family of functions $$f_{a,b} : [0,1] \to [0,1]$$ of the form $$f_{a,b}(x) = (1 - x^a)^{1/b}$$ where $a, b$ are positive even integers.  Then note that the inverse function of $f_{a,b}$ is $f_{b,a}$.  Since both functions are nonnegative and monotone decreasing, it immediately follows that $$\int_{x=0}^1 f_{a,b}(x) - f_{b,a}(x) \, dx = 0$$ for all such positive even integers $a, b$.  In fact the identity holds for any positive reals $a, b$ but the positive even case is easy to see and is applicable to your case.