The problem is to show the following:
Let $\varphi$ be a closed concave function, and $M=\max_{x \in \mathbb{R}^d} \varphi(x)$. Let $D_r:=\{\varphi\geq r\}$ be the level set. Then given $r \leq s \leq M$, $$\lambda_d(D_s)\geq \bigg(\frac{M-s}{M-r}\bigg)^d \lambda_d(D_r)$$ where $\lambda_d$ is the Lebesgue measure on $\mathbb{R}^d$.
The intuition is quite clear, but I'm not sure how to proceed with that. Any comment shall be greatly appreciated.
Hint. First think $d = 2$ (or $3$). Consider a triangle (cone) with base $D_r$ and height $M-r$. Show that for any horizontal slice at height $s-r$, the legth (area) of that slice equals $\frac {M-s}{M-r}D_r$.