We need to show a couple of lemmas to solve this problem. Lemma 1: Let $b, c, \beta$ and $\gamma$ be arbitrary positive real numbers. Then, we have \[\frac{b\beta}{1+\beta} + \frac{c \gamma }{1+ \gamma} \geq \frac{\text{min}(b\beta+c\gamma, (b+c)\beta\gamma)}{\beta + \gamma}.\] Proof: The first step is to write $\beta' = 1/\beta$ and $\gamma' = 1/\gamma$. The inequality is now equivalent to \[\frac{b}{1 + \beta'}+\frac{c}{1 + \gamma'} \geq \frac{\text{min}(b \gamma' + c \beta', b+c)}{\beta'+\gamma'}\]We will break into two cases here. Case 1. $b+c \geq b\gamma' + c\beta'$ In this case, our required inequality is \[\frac{b}{1+\beta'}+\frac{c}{1+\gamma'} \geq \frac{b\gamma'+c\beta'}{\beta'+\gamma'}\]Expressing $T = b\gamma'+c\beta'$ the above becomes \[\frac{b+c+T}{1+\beta'+\gamma'+\beta'\gamma'} \geq \frac{T}{\beta'+\gamma'}\]Post cross multiplication, it suffices to prove \[(b+c)(\beta'+\gamma') + T(\beta' + \gamma') \geq T(1+\beta'+\gamma'+\beta'\gamma')\]Remarking that \[(b+c)(\beta'+\gamma') = T+b\beta'+c\gamma'\]the above finally reduces to \[b\beta'+c\gamma' \geq T\beta'\gamma'\]Dividing throughout by $\beta'\gamma'$ it suffices to show \[\frac{b}{\gamma'}+\frac{c}{\beta'} \geq b \gamma' + c \beta'\]The key estimate now comes from Titu's lemma: \[\frac{b}{\gamma'}+\frac{c}{\beta'} = \frac{b^2}{b \gamma'} + \frac{c^2}{c \beta'} \geq \frac{(b+c)^2}{b \gamma' + c \beta'} \geq b \gamma' + c \beta'\]where the last inequality follows from $b+c \geq b \gamma' + c \beta'$. Case 2: $b\gamma' + c\beta' > b+c$. In this case we must show \[\frac{b}{1 + \beta'}+\frac{c}{1 + \gamma'} \geq \frac{b+c}{\beta'+\gamma'}\]We use the variable $T$ from Case 1. The above inequality is equivalent to \[\frac{b+c+T}{1+\beta'+\gamma'+\beta'\gamma'} \geq \frac{b+c}{\beta'+\gamma'}\]Cross multiplying and canceling the common term $(b+c)(\beta'+\gamma')$, the above is seen to be the same as \[T(\beta'+\gamma') \geq (b+c)(1+\beta'\gamma')\]Expressing $T = b\gamma' + c\beta'$ and canceling $(b+c)\beta'\gamma'$ from both sides, this finally becomes equivalent to showing \[b\gamma'^2+c\beta'^2 \geq b+c\]Fortunately, this is again proven from Cauchy Schwarz, since one has \[(b\gamma'^2+c\beta'^2)(b+c) \geq (b \gamma' + c\beta')^2 \geq (b+c)^2\]Combining Cases 1 and 2, the original estimate is proven. $\square$ Lemma 2: Let $ABC$ be a triangle with side lengths $a, b, c$ defined as usual. Let $\beta, \gamma$ denote arbitrary positive real numbers. If there exists a point $P$ such that \[PA = PB \gamma = PC \beta\]then we have \[b \beta + c\gamma \geq a \beta \gamma\]Proof: We have from Ptolemy's inequality: \[PB \cdot CA + PC \cdot AB \geq PA \cdot BC\]Rearranging and using the ratios, it can be seen this is equivalent to the required equality. With the above two estimates, we are ready to destroy the problem. Let $A, B, C, D$ be any four points in the plane. In what follows, we shall assume that no three among $A, B, C, D$ are collinear. The other configurations can be settled using geometric continuity. Let $\beta, \gamma$ be the positive real numbers such that \[DA = DB \gamma = DC \beta \qquad (1)\]We first recentre our inequality from $B$ to $A$, so that we are required to show \[\frac{c}{DA+DB}+\frac{b}{DA+DC} \geq \frac{a}{DB+DC}\]Multiplying throughout by $DA$ and using $(1)$ the above becomes \[\frac{c \gamma}{1+ \gamma} + \frac{b \beta}{1 + \beta} \geq \frac{a \beta \gamma}{\beta + \gamma} \qquad (2)\]Now from Lemma 1 we have \[\frac{b\beta}{1+\beta} + \frac{c \gamma }{1+ \gamma} \geq \frac{\text{min}(b\beta+c\gamma, (b+c)\beta\gamma)}{\beta + \gamma}\]and therefore, it suffices to show that \[\text{min}(b \beta + c\gamma, (b+c)\beta\gamma) \geq a \beta \gamma\]But this is clear, since from Lemma 2 we have \[b \beta + c \gamma \geq a \beta \gamma\]and \[(b+c)\beta \gamma \geq a \beta \gamma\]from the triangle inequality in $\triangle ABC {}$. This shows that $(2)$ is indeed true, and the problem is solved. $\square$

The cosine law part in the proof of Lemma 2 seems unnecessary to me, and can probably be replaced with a smarter (possibly geometrical) argument. Unfortunately, I couldn't spot such an argument. Another thought: the geometric continuity argument can probably be avoided by explictly handling those configurations. EDIT: The cosine law part is indeed bogus. It turns out I, being the genius that I am, proved Ptolemy's inequality using Apollonian circles. Completely pointless, and I still don't know what I was thinking. Here's the proof if you're interested: We'll be working with signed distances on lines $AB, AC$ henceforth, with the positive direction being along $A \mapsto B$ and $A \mapsto C$ respectively. Let $Y$ denote the unique point on segment $AC$ such that $AY = \beta YC$. Define $Z$ similarily with $AZ = \gamma ZB$. Consider $\omega_B$, the Apollonian circle of segment $AC$ passing through $Y$. Define $\omega_C$ similarily. The existence of such a point $P$ implies that $\omega_B, \omega_C$ have a common point. Letting $r_B, r_C$ denote the radii of $\omega_B, \omega_C$ and $O_B, O_C$ their centres, we obtain that \[|r_B-r_C| \leq O_BO_C \leq r_B+r_C\]which after squaring and rearrangement takes the form \[|r_B^2 + r_C^2 - O_BO_C^2| \leq 2r_Br_C \qquad (1)\]We intend on computing each of the above distances. Let us compute $r_B$ first. We know that $O_BA = \beta^2 \O_BC$ which implies that \[O_BA = \frac{\beta^2 b}{1-\beta^2}; O_BC = \frac{\beta^2}{1-\beta^2} \qquad (2)\]It follows from $r_B^2 = O_BA \cdot O_BC$ that $r_B = \frac{\beta b}{|1-\beta^2|}$. Similarily, one has $r_C = \frac{\gamma c}{|1-\gamma^2|}$. The analogous first equality of $(2)$ also gives \[O_CA = \frac{\gamma^2 c}{1-\gamma^2}\]From the Cosine Law in triangle $AO_BO_C$ (remember we're working in signed distances, so the angle used is simply $\angle BAC$) we now obtain \[O_BO_C^2 = O_BA^2 + O_CA^2 - 2O_BO_C\cos A = \frac{\beta^2 b^2}{(1-\beta^2)^2} + \frac{\gamma^2 c^2}{(1-\gamma^2)^2} - \frac{(2bc\cos A) \beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)}\]Plugging this in $(1)$ we have \[\left| \left(r_B^2 - \frac{\beta^4 b^2}{(1-\beta^2)^2}\right) + \left(r_C^2 - \frac{\gamma^4 c^2}{(1-\gamma^2)^2}\right) + \frac{(2bc\cos A) \beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)} \right | \leq 2r_Br_C \qquad (3)\]Note that \[\left(r_B^2 - \frac{\beta^4 b^2}{(1-\beta^2)^2}\right) = \left(\frac{\beta^2b^2}{(1-\beta^2)^2} - \frac{\beta^4 b^2}{(1-\beta^2)^2}\right) = \frac{\beta^2 b^2}{1-\beta^2}\]Plugging in this much required simplification and it's $C$-analogue in $(3)$ we obtain \[\left| \frac{\beta^2 b^2}{1-\beta^2} + \frac{\gamma^2 c^2}{1-\gamma^2} + \frac{(2bc\cos A) \beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)} \right | \leq 2r_Br_C\]We finally unleash $2bc \cos A = b^2+c^2-a^2$ in the above estimate to obtain \[\left| \left(\frac{\beta^2 b^2}{1-\beta^2} + \frac{b^2\beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)}\right) + \left(\frac{\gamma^2 c^2}{1-\gamma^2} + \frac{c^2\beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)}\right) - \frac{a^2\beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)} \right | \leq 2r_Br_C\]The first bracketed term above is actually just \[\frac{\beta^2 b^2}{1-\beta^2} + \frac{b^2\beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)} = \frac{\beta^2b^2}{(1-\beta^2)(1-\gamma^2)}\]Therefore our earlier estimate takes on the much more tame form: \[\left|\frac{\beta^2 b^2 + \gamma^2 c^2 - a^2\beta^2 \gamma^2}{(1-\beta^2)(1-\gamma^2)}\right| \leq 2r_Br_C \qquad (4)\]Using the fact that $r_B|1-\beta^2| = \beta b$ (and it's $C$-analogue) we can further simplfy the above to \[|\beta^2 b^2 + \gamma^2 c^2 - a^2\beta^2 \gamma^2| \leq 2 bc \beta \gamma\]This implies that $a^2\beta^2\gamma^2$ lies between \[\beta^2b^2 + \gamma^2c^2 - 2bc\beta \gamma = (b \beta- c \gamma)^2 \text{ and } \beta^2 b^2 + \gamma^2 c^2 + 2bc \beta \gamma = (b \beta + c \gamma)^2\]We therefore obtain \[b \beta + c \gamma \geq a \beta \gamma \geq |b \beta - c \gamma|.\]The left estimate above is precisely what we wanted to show, and therefore we are done. $\square$