Q:

Metric/bilinear form triangle inequality proof

Let V be a vector space and $g(u,v)$ be any symmetric, positive definite, non degenerate bilinear form defining a Euclidean scalar product in V. Define the distance function;
$$d(u,v)=g(u-v,u-v)^{\frac{1}{2}}$$
Prove that $d(u,v)$ satisfies the triangle inequality. 
As intermediate steps it asks to prove the following (considering $d^2(u,v)=g(u-v,u-v)$);
a) $d(u,0)\leq d(u,v)+d(v,0)$
b) $d(u,v)=d(u+w,u+w)$
I have proved both of the above but I don't see how a) comes into use when proving the triangle inequality.  
I have tried the following proof, but there must be something wrong with it since it is so simple and doesn't use a):
$d(u,w)=d(u+v,w+v)$
$~~~~~~~~~~~~=d(u,v)+d(v,w)+d(v,v)+d(u,w)$
$d(v,v)+d(u,w)\geq 0 \implies d(u,w)\leq d(u,v)+d(v,w)$
Can someone point out where my false step is, and how I am to incorporate a) into my proof?  Thank you.  

A:

Your second equality (expanding out $d$) is actually for $d^2$, which is bilinear; $d$ itself isn't. 

What it probably wants is
$$ d(u,v) = d(u-v,0) \leq d(u-v,w-v) + d(w-v,0) = d(u,w)+d(w,v), $$
using property (a), then (b), then (a) again.