An inequality for $2$-convex and concave functions

Let $I$ be an interval of the real line and $f:I\rightarrow \mathbb{R}$ a function. Define the divided difference of $f$ at the points $x_0<x_1<\cdots<x_{n+1}$ in $I$ by $$[x_0;f]=f(x_0)\; , \; [x_0,x_1,\cdots,x_{k};f]=\frac{[x_1,\cdots,x_{k};f]-[x_0,\cdots,x_{k-1};f]}{x_k-x_0},$$ for every $1\leq k\leq n+1$. The function $f$ is called convex of order $n$ or $n$-convex if for any system $x_0<x_1<\cdots<x_{n+1}$ of points in $I$ it holds that $[x_0,x_1,\cdots,x_{n+1};f]\geq 0$, and it is $n$-concave if $-f$ is $n$-convex.

Now, can somebody prove that:

(a) If $f:\mathbb{R}^+\rightarrow \mathbb{R}^+$ is a function such that $\log f$ is concave of order $2$, then for every $0<u<v<w<z$ the following inequality holds \begin{equation} \left(\frac{f(z)}{f(u)}\right)^{w-v}\leq \left(\frac{f(w)}{f(v)}\right)^{z-u} \end{equation}

(b) If $f$ is three times differentiable on $I$, then $f$ is convex of order two if and only if $f'''(x)\geq 0$, for all $x\in I$.