First we show $d(X,AB)+d(X,AC)=d(X,BC)$. This is true when $X\equiv E$ or $F$, so it is true for any $X$ on $EF$ because each of the terms vary linearly when we move $X$ along the line $EF$ linearly. So now $d(X,AB)=d(X,BC)-d(X,AC)\leq XZ+d(Z,BC)-d(X,AC)$, and summing all these up we get the desired result.

oneplusone wrote: First we show $d(X,AB)+d(X,AC)=d(X,BC)$. This is true when $X\equiv E$ or $F$, so it is true for any $X$ on $EF$ because each of the terms vary linearly when we move $X$ along the line $EF$ linearly. Can someone give more explanations why moving $X$ along the line $EF$ keeps the equality $d(X,AB)+d(X,AC)=d(X,BC)$?

Well the condition describes a line in trilinears ,line is spanned by two points and so it's enough to show that for two points. Quite a common idea that works wonders for higher order curves,but if you just wanna' dumb it down put in a matrix and don't think about it ,you don't even need them to be normalised.