we have: A function $f:I\rightarrow\mathbb{R}^+$ is log-convex if and only if \begin{align*} f(\lambda x + u y)\leq f^{\lambda}(x) f^u (y) \end{align*} for $x, y\in I$ and $\lambda, u>0$ with $\lambda + u = 1$.
Now, my question is: why? The reformulation of log-convexity implied by above inequality is equivalent to the following working definition: the function $f:I\rightarrow\mathbb{R}$ is log-convex if and only if for all $x, y, z\in I$ with $x<y<z$ \begin{align*} f^{z-x}(y)\leq f^{z-y}(x) f^{y-x}(z). \end{align*}
The two are equivalent. If $x <y<z$ we can write $y=\lambda z+(1-\lambda)x$ where $\lambda=\frac {y-x} {z-x}$. Applying the first definition to this we get the second one. [ We get $f(y) \leq f(z)^{\frac {y-x} {z-x}} f(x)^{(1-\frac {y-x} {z-x})}$. Rise to power $z-x$]. Conversely, give, $x<z$ and $\lambda ,u >0$ with $\lambda +u =1$ take $y=\lambda z+(1-\lambda)x$. The $x <y<z$ and we can use the second definition to get the first (with $z$ in place of $y$).