I'm doing a paper work about theory of means and their inequalities and I'm trying to prove that if A denotes the arithmetic mean and I the identric mean then $$ I(x, y) \le A(x, y) $$ for $x,y>0$, i.e. \begin{equation} \frac{1}{e}\left(\frac{y^y}{x^x}\right)^{\frac1{y-x}} \leq\frac{x+y}{2} \, . \end{equation}
Every article I have researched send me to this one: https://www--tandfonline--com.uma.debiblio.com/doi/abs/10.1080/0025570X.1975.11976447 where it is supposed to be proved, but that has to be payed for.
We must assume that $x \ne y$ because the left-hand side is not defined for $x=y$. Then we can assume that $x<y$ because both sides not change if $x$ and $y$ are exchanged.
Taking logarithms, we need to show that for $0 < x < y$ $$ \frac{y \ln y - x\ln x}{y-x} - 1 \le \ln \left( \frac{x+y}{2}\right) \, . $$ Using an idea from this answer to a related question we write the left-hand side as an integral: $$ \frac{1}{y-x}\int_x^y \ln(t) \, dt \le \ln \left( \frac{x+y}{2}\right) \, . $$ And this is true because the logarithm is a concave function.