Monotonicity of special case of Kullback-Leibler divergence I have two discrete distributions $\tau$ and $\rho$ with the same support $\Omega$. I'm considering a weighted mixture of these distributions described by the following function:
$$
f(w) = (1-w) \cdot \tau + w \cdot \rho, ~~ \text{where} ~~ w \in [0,1]
$$
I'm particulary interested in very special case of Kullback-Leibler divergence / relative entropy:
$$
KL(f(w), \rho) = \sum\limits_{i \in \Omega}
((1-w) \cdot \tau_i + w \cdot \rho_i) \cdot
\ln (\frac{(1-w) \cdot \tau_i + w \cdot \rho_i}{\rho_i})
$$
Generally, $KL(f(0), \rho) \geq 0$ and decreases towards $KL(f(1), \rho) = 0$ as $w$ goes from $0$ to $1$.
I would like to formally assure that this very specific case of Kullback-Leibler divergence is MONOTONICALLY decreasing, because I wonder if there's a possibility of such phenomenon that I would call "cannot escape from the valley", which is depicted below:

 A: According William A. Huber comment, which solves the problem and proves that this function is monotonically decreasing:
The second derivative is always non-negative.
$$
h_i''(w) =
\frac{(\rho_i - \tau_i)^2}{\rho_i} \cdot
\frac{1}{(1 - w) \cdot \frac{\tau_i}{\rho_i} + w}
$$
$$
(1 - w) \cdot \frac{\tau_i}{\rho_i} + w =
(1 - \frac{\tau_i}{\rho_i}) \cdot w + \frac{\tau_i}{\rho_i}
~~\text{this is linear function of $w$}
$$
This linear function from denominator takes values from
$[\frac{\tau_i}{\rho_i}, 1]$ or $[1, \frac{\tau_i}{\rho_i}]$, thus
the whole second derivative is non-negative, hence convex and I won't get such "valley".
As William cleverly pointed out, that's no the end, because I still must prove that this function is decreasing. I can accomplish that by using $h_i$ convexity and Jensen's inequality:
$$\forall_{t \in [0,1]}~
h(t \cdot w_1 + (1-w) \cdot w_2) \leq t \cdot h(w_1) + (1-t) \cdot h(w_2)
$$
Let's fix $w_2 = 1$ for which $h(w_2) = h(1) = 0$.
Now, $w_3 = t \cdot w_1 + (1-w)$ denotes some point to the right of
$w_1$ (between $w_1$ and $1$):
$$\forall_{t \in [0,1]}~
h(t \cdot w_1 + (1-w) \cdot 1) \leq t \cdot h(w_1)
$$
So, we have $w_1 \leq w_3$ and thankfully $h(w_1) \geq h(w_3)$, thus
this function is monotonically decreasing.
