Let $\beta\geq1$, $n\in\mathbb N$ and $r_1,\dots, r_n\geq1-\frac1\beta$ be real numbers.
I want to prove that
Claim. $$\prod_{k=1}^n (1+\beta(r_k-1))\le\left(1+\beta\left(\sqrt[n]{\prod_{k=1}^n r_k}-1\right)\right)^n.$$
My thoughts.
It can be rewritten as $$\prod_{k=1}^n ((1-\beta)+\beta r_k)\le\left((1-\beta)+\beta\sqrt[n]{\prod_{k=1}^n r_k}\right)^n,$$
which looks like generalized Hölder with the wrong direction, however, since $\beta\geq1$, we have $1-\beta\le0$ and thus classical Hölder doesn't apply.
What could be done to prove this?
Note: Using induction, I can reduce the problem to showing that for all $\beta\geq1, n\in\mathbb N, R\geq\left(1-\frac1\beta\right)^n,r\geq1-\frac1\beta$,
$$((1-\beta)+\beta r) ((1-\beta)+\beta\sqrt[n]R)^n\le((1-\beta)+\beta\ \sqrt[n+1]{Rr})^{n+1},$$
which once again looks like the classical version of Hölder with wrong direction (but again, $1-\beta\le0$).
Notice that the function $f(x)\overset{\text{Def.}}=\log((1-\beta)+\beta e^x)$ is concave since $f''(x)=\frac{\beta(1-\beta)e^x}{((1-\beta)+\beta e^x)^2}\le0$.
By Jensen's inequality, we have $$\log(\text{LHS})=\sum\limits_{i=1}^n f(\log r_i)\le n f\left(\frac{\sum_{i=1}^n\log r_i}n\right)=\log(\text{RHS}).\quad \blacksquare$$