Inequality from a point in plane to a triangle OR Inequality on a quadrilateral 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
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?.
 A: 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 ${\mathbb R}^2$.  
A: As we see from the solution by Iosif Pinelis, triangle inequality for $ABC$ and Ptolemy inequality for $A,B,C,D$ are enough. In particular, this holds in $\mathbb{R}^3$ and in any Ptolemaic space. 
Here is bit simplified argument of Iosif Pinelis. Denote $AB=\gamma$, $BC=\alpha$, $AC=\beta$, $DA=a$, $DB=b$, $DC=c$. We have (1) and (2) and want (3): 
\begin{align}
\alpha+\gamma\geqslant \beta\,\,\,\,(1)\\
\frac ab\alpha+\frac cb\gamma\geqslant \beta \,\,\,\,(2)\\
\frac {a+c}{b+c}\alpha+\frac {a+c}{a+b}\gamma\geqslant \beta.\,\,\,\,(3)
\end{align}
If $b\leqslant \min (a,c)$, (3) follows from (1). If $b\geqslant \max (a,c)$, (3) follows from (2) (since $(a+c)/(c+b)\geqslant a/b$.) Finally, if $b$ lies between $a$ and $c$, multiplying (1) and (2) by the factor $\frac{b(a+c)}{(b+a)(b+c)}$ and summing up we get 
$$
\frac {a+c}{b+c}\alpha+\frac {a+c}{a+b}\gamma\geqslant \frac{2b(a+c)}{(b+a)(b+c)}\beta\geqslant \beta
$$
as desired.
