Define the multi-dimension concave function $f(x): \mathbb{R}^n_+ \rightarrow \mathbb{R}_+$ where $x \in \mathbb{R}^n_+$, here I use $\mathbb{R}_+$ to represent the range $[0, \infty)$ and we let $f(\mathbf{0}) = 0$ . Suppose we have a vector $x^* = [1, ..., 1]$, let $x^*_{-i} = [1, ..., 0, ..., 1]$ whose $i^{th}$ element is 0. Note the value of $x^*$ can be any positive value, I just use $1$ for convenience.
Let $a = f(x^*)$ and $b = \sum\limits_{i=1}^{n} f(x^*_{-i})$, I have an intuition that \begin{equation} (n - 1) \cdot a \geq b \end{equation} because of the concavity of $f(x)$. But I cannot think out a proof for this. Does anyone have some thoughts about doing this or prove my intuition is wrong?