Inequality from a point in plane to a triangle OR Inequality on a quadrilateral

Question

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?.

Answer 1

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 ${\mathbb R}^2$.

Answer 2

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.