Let $E$ & $F$ be points such that $BCDE$ and $BCAF$ are both parallelograms.Then $EDAF$ is also a parallelogram. Hence, $AF=ED=BC,EF=AD,EB=CD,BF=AC$.Now applying ptolemy's inequality to $ABEF$ and $AEBD$ we have $AB.AD+BC.CD=AB.EF+AF.BE\ge AE.BF=AE.AC$ and $BD.AE+AD.CD=BD.AE+AD.BE\ge AB.ED=AB.BC$. Therefore $DA.DB.AB+DB.DC.BC+DC.DA.CA=DB(DA.AB+DC.BC)+DC.DA.CA\ge DB.AE.AC+DC.DA.CA=AC(DB.AE+DC.DA)\ge AC.AB.BC$ Now the given problem is the equality case which occurs iff $ABEF$ as well as $AEBD$ are cyclic, i.e $AFED$ is a parallelogram it follows that $AD\perp ED$ i.e $AD\perp BC$ similarly we get that $CD\perp AB$.Thus $D$ is the orthocenter of $ABC$.

Hello, does there exist a solution involving an inversion centered at $D$?

I would like to mention, that problem 4428 from Crux Mathematicorum by Leonard Giugiuc can be solved using the same trick. For all complex numbers $u,v,w$ holds $$|uv(u-v)|+|vw(v-w)|+|wu(w-u)|\ge |(u-v)(v-w)(w-u)|$$Just substitute $$u=b+c,\ v=c+a,\ w=a+b$$to get the key estimation as in solution I. References: Crux Mahematicorum vol. 45(8)

joybangla wrote:

Use rotation property. We get $\measuredangle ABC=\measuredangle CDA$. Then it can be seen that $D$ is orthocenter.