If points $A$, $B$, $C$ form a triangle in euclidean space and $D$ is another point in the plane of the triangle, the problem is to show that : 


$\frac{AB}{DA + DB} + \frac{BC}{DB + DC} \ge \frac{AC}{DA + DC}$


[Figure here](https://i.stack.imgur.com/IWKZj.png)


I verified that it holds for more than several million positions of point $D$, including when point $D$ is inside triangle $ABC$. However, I can't seem to give a proof for it.
The only inequality that resembled the one above was the Ptolemy's theorem which gives a relation between the four sides of a quadilateral made from points $A,B,C, D$ and two diagonals $AC,BD$:


$AB.DC + BC.DA \ge AC.DB$ which in other terms is $\frac{AB}{DA \cdot DB} + \frac{BC}{DB \cdot DC} \ge \frac{AC}{DA \cdot DC}$


This has multiplication in the denominator but what I am looking for is addition. Any ideas on how to go about it or perhaps some inequality results that can be applied to prove it ? Or do you see a special instance wherein the inequality doesn't hold?.



Inequality from a point in plane to a triangle OR Inequality on a quadrilateral
I would like to streamline the proof a bit. 


If $AC=0$ then the inequality in question is obvious. So, assume that $AC\ne0$. Multiplying both sides of the inequality by $(AD + BD) (AD + CD) (BD + CD)$, rewrite it as 
$$
 L:=(AD + CD) (AD\ BC + AB\ BD + BC\ BD + AB\ CD)
 -AC (AD + BD) (BD + CD)\ge0. 
$$
Clearly, $L$ is concave in $BD$. 
So, in view of Ptolemy's inequality $AB\ CD + BC\ AD\ge AC\ BD$, without loss of generality either $BD=0$ or $BD=(AB\ CD + BC\ AD)/AC$. 
But, by the triangle inequality $AB + BC \ge AC$,  

$$L\big|\_{BD=0}=AD^2 BC + AD (AB + BC - AC) CD + AB\ CD^2\ge0
$$
and 
$$
 AC\ L\big|\_{BD=(AB\ CD + BC\ AD)/AC} 
=\big((AB + BC)^2 - AC^2\big) AD\ CD + AB\ BC (AD - CD)^2\ge0. 
$$
This completes the proof. 


In conclusion, one may note that inequality in question may fail to hold in some metric spaces $M$, even if $M$ is a normed space. Let e.g. $M={\mathbb R}^2$ with the $\ell^1\_2$ norm $\|(x,y)\|:=|x|+|y|$ for $(x,y)\in{\mathbb R}^2$, and then let $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$, so that 
$AB=BC=AD=CD=1$ and $AC=BD=2$. Here $XY:=\|X-Y\|$ for any $X$ and $Y$ in ${\mathbb R}^2$. 


