Let us start with an example, consider the following sum: $$H_n=\sum_{i=1}^{n}\frac{1}{n-1+x_i}$$ With $x_1x_2...x_n=1$, We're willing to show that $H_n \le 1$.
To do this we start with the following lemma:
Lemma: If a differentiable function has a single inflexion point and is evaluated at n arbitrary reals with a fixed sum, any minimum or maximum must occur where some $n − 1$ variables are equal.
Proof [From mildorf introduction to inequalities]: Consider $f(y)=\frac{1}{k+e^y}$ for a non-negative constant $k$. We have $f''(y)=\frac{e^y(e^y-k)}{(k+e^y)^3}$, so $f''(y)\ge 0 \iff e^y\ge k$, thus $f(y)$ has exactly one inflexion point at $y=\ln{k}$ and $f$ is convex in the interval $(\ln k,\infty) $. Consider now $y_i=\ln{(x_i)}$ therefore $y_1+y_2+...y_n=0$, and if we take $k=n-1$ we have:$$H_n=\sum_{i=1}^{n}f(y_i)$$ The inflection point will be at $k_0=\ln(k)=\ln(n-1)$ Suppose $y_1 \ge y_2 \ge...\ge y_m \ge k_0 \ge y_{m+1}... \ge y_n $ for some positive $m$. Then by Karamata's Inequality (also called majorization inequality):$$f(y_1)+f(y_2)+...f(y_m) \le (m-1)f(k_0)+f(y_1+...+y_m-(m-1)k_0) \:\:\:[1] $$ Also by Karamata: $$(m-1)f(k_0)+f(y_{m+1})+...+f(y_n) \le (n-1)f(\frac{(m-1)k_0+y_{m+1}+...+y_n}{n-1}) \:\:\: [2]$$ "otherwise, all of the $y_i$ are less than $k_0$ In that case we may directly apply Majorization to equate $n-1$ of the $y_i$ whilst increasing the sum in question. Hence, the lemma is valid"
With this lemma we can prove our inequality quite easly, so I'll skip on that. My question is, in the proof I understood everything up to the very end, I've put quotation marks on the part that confused me, I really don't see his arguement or what he is trying to say in those few lines and I'm kinda lost on why does this prove our initial statement, any help is appreciated!