Proving $N(x) = (A(x,x))^\frac12$ is a norm in $V$ given that A is a positive definite bilinear form The problem: Given a vector space $V$ with scalars in $\mathbb{R}$, and a positive-definite bilinear form $A$, show $N(v)=(A(v,v))^\frac12$ defines a norm on $V$.
I have easily handled everything but the inequality requirement:
$N(u + v) \leq N(u) + N(v)$
From searching around I have found the hint to use the Cauchy-Schwarz inequality:
$(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$
But I still don't see it.
If I take $A$ to be a dot product, an example of a function that satisfies the definition, I can finally do it:
$N(u + v)^2 = \sum (u_i + v_i)^2 = \sum u_i^2 + 2u_i v_i + v_i^2$
$(N(u) + N(v))^2 = N(u)^2 + 2 N(u) N(v) + N(v)^2 = \sum u_i^2 + 2 \sqrt{\sum u_i^2 \sum v_i^2} + \sum v_i^2$
Subtracting and dividing out the common parts and squaring it fits right into Cauchy-Schwarz.
However I don't know how to do this if I can't assume $A(u,v) = u \cdot v$
 A: typical approach: since you've proven positive definiteness, square each side and prove the equivalent claim
$ \big(N(u + v)\big)^2 $
$=A(u+v, u+v) $
$=A(u,u) + A(v,v) + A(u,v)+ A(v,u)$
$=A(u,u) + A(v,v) +2\Big(\frac{1}{2}\big(A(u,v)+ A(v,u)\big)\Big)$
$\leq  A(u,u) + A(v,v) +  2\cdot A(u,u)^\frac{1}{2}A(v,v)^\frac{1}{2}$
$=\big(N(u) + N(v)\big)^2 $
which holds by Cauchy-Schwarz
addendum
for a short proof of Cauchy Schwarz, since you've proven positive definiteness of the bilinear form, consider for any two vectors $v,u\in V$ the candidate Gram matrix
$G:=\displaystyle \left[\begin{matrix}A( v, v) & A(v, u)\\ A( u, v)& A( u, u)\end{matrix}\right]$
Positive definiteness of the bilinear form implies for any $\mathbf x\in \mathbb R^2$
$\mathbf x^T G \mathbf x =A\Big(( x_1v+x_2 u), (x_1v + x_2 u)\Big) \geq 0$
with $G':= \Big(\frac{1}{2}\big(G+G^T\big)\Big)$ it also implies $\mathbf x^T G' \mathbf x \geq 0$
note that $G'$ is real symmetric and hence a proper Gram matrix.  Now $G'\succeq 0\implies \text{eigs}\geq 0$ which means its determinant is $\geq 0$, thus
$A( v, v)\cdot A( u, u) - \Big(\frac{1}{2}\big(A(u,v)+A(v, u)\big)\Big)^2 =\det\big(G'\big) \geq 0 $
which simplifies to
$  \frac{1}{2}\big( A(u,v)+A(v, u)\big)  \leq A( u, u)^\frac{1}{2}\cdot A( v, v)^\frac{1}{2}$
as used in the proof of triangle inequality
A: Let $B=\left \{ e_1,\dots,e_n \right \}$ be a basis of $V$ and denote the matrix $\tilde{A}=[A(e_i,e_j)]_{ij}$. Then we have that for all $u,v\in V$ by denoting their components as $[u]_B = [u_1\ \cdots \ u_n ]^T$ using bilinearity
$$A(u,v)=\sum_{i,j=1}^n A(e_i,e_j) u_i v_j=[u]_B^T\tilde{A}[v]_B$$
This helps us to basically reduce to the $\mathbb R^n$ case using matrix language. Since $A$ is PD then so is $\tilde{A}$ as a matrix. Resorting to the spectral decomposition $\tilde{A}=Q\Lambda Q^{T}$ for $Q$ unitary and $\Lambda$ diagonal with postivie entries, we get by denoting $v'=Q^{T}[v]_B$:
$$A(u,v)=[u]_B^TQ\Lambda Q^{T}[v]_B=u'^T\Lambda v'=\sum_{i=1}^n \lambda_i u_i' v_i'$$
Let's compute then
$$N(u+v)^2=(u+v)'^T\Lambda (u+v)=(u'+v')^T\Lambda(u'+v')=N(u)^2+N(v)^2+2u'^T\Lambda v'$$
if we were to show $u'^T\Lambda v'\leq N(u) N(v)$ we would be done by simple algebra, since then $N(u)^2+N(v)^2+2u'^T\Lambda v'\leq (N(u)+N(v))^2$.
This can be proven using the basic Cauchy-Schwarz
$$u'^T\Lambda v'= \sum_{i=1}^n\lambda_i u_i' v_i' =  \sum_{i=1}^n\lambda_i^{\frac{1}{2}}u_i' \cdot\lambda_i^{\frac{1}{2}}v_i'\leq \left ( \sum_{i=1}^n \left ( \lambda_i^{\frac{1}{2}}u_i' \right )^2 \right )^{\frac{1}{2}} \left ( \sum_{i=1}^n \left ( \lambda_i^{\frac{1}{2}}v_i' \right )^2 \right )^{\frac{1}{2}} = \left ( \sum_{i=1}^n  \lambda_i u_i'^2  \right )^{\frac{1}{2}}\left ( \sum_{i=1}^n  \lambda_i v_i'^2  \right )^{\frac{1}{2}}=N(u)N(v)$$
