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)
Best Answer
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.