Trace of a Matrix is positive I wish to prove that the trace of a matrix is positive. 
The matrix is defined as follows 
$B_+ = (B - \frac{Bxx^TB}{x^TBx})$ where B is symmetric positive definite. 
Now Ive done the following:
$tra(B_+) = tra(B - \frac{Bxx^TB}{x^TBx}) = tra(B) - tra(\frac{Bxx^TB}{x^TBx}) = tra(B) - tra(\frac{(Bx)^TBx}{x^TBx}) = tra(B) - (\frac{||Bx||}{x^TBx}) > tra(B) - \frac{||B|| ||x||}{x^TBx}$
This is as far is Ive gotten. Is there anything Im missing to get to tra(B)>0
 A: as stated, with $B\succeq 0$ there are some problems that break the OP's goal, e.g. with dividing by zero when considering $\mathbf x^T B \mathbf x=0$ when $\mathbf x$ is in the nullspace of $B$.  Also it technically isn't true that $tr(B^+)\gt 0$, only that $tr(B^+)\geq 0$
(For example, consider the cases when $B$ has rank one or rank zero.)    
I prove the result assuming that $B\succ 0$
by application of matrix determinant lemma (for rank one updates):
$\det\big(B^+\big) = \det\big(B\big)\big(1+ \frac{-1}{\mathbf x^TB \mathbf x}(\mathbf x^TB)B^{-1}(B\mathbf x))$ $= \det\big(B\big)\big(1+ \frac{-1}{\mathbf x^TB \mathbf x}\mathbf x^TB\mathbf x)) = \det\big(B\big)\big(1-1\big)=0$ 
so $B^+$ is necessarily singular  
to cleanup notation
$\mathbf y: =  \frac{1}{\sqrt{\mathbf x^TB \mathbf x}}B\mathbf x$
$B^+ = B - \mathbf{yy}^T$ 
when we look at a quadratic form, for nonzero $\mathbf z \perp \mathbf y$
$\mathbf z^T B^+\mathbf z = \mathbf z^T B\mathbf z - \mathbf z^T\mathbf{yy}^T\mathbf z = \mathbf z^T B\mathbf z +0\gt 0$
by positive definiteness of $B$.
$\mathbf z$ lives in a $n-1$ dimensional subspace, and we know $B^+$ is singular, so it follows that $\mathbf y$ is in the kernel of $B^{+}$ and $B^+$ has signature of $(n-1,0,1)$, i.e. $B^+$ is a non-zero positive semi-definite matrix which implies $tr\big(B^+\big) \gt 0$ 
addendum
here's an analytic proof
consider the path
$B(\tau) = (1-\tau)B+ \tau B^{+} = B- \tau \mathbf{yy}^T$
for $\tau \in [0,1]$ 
now re-visit the matrix determinant lemma for rank one updates to get
$\det\big(B(\tau)\big)= \det\big(B\big)\big(1+ \tau\frac{-1}{\mathbf x^TB \mathbf x}\mathbf x^TB\mathbf x))= \det\big(B\big) \cdot (1-\tau)$
which is nonzero for $\tau\in[0,1)$ 
by topological continuity of eigenvalues this means that the eigenvalues of $B(\tau)$ are all positive for $\tau \in [0,1)$ i.e. $B$ has all positive eigenvalues because it is real symmetric positive definite -- and, essentially, apply intermediate value theorem -- so no eigenvalue can have 'crossed over' to negative without making $B(\tau)$ singular for some $\tau\in[0,1)$.  (A more careful way to do this is with winding numbers but I digress.)      
Further applying continuity, this means at $\tau =1$, $B(\tau) = B^{+}$ cannot have negative eigenvalues and since it is real symmetric this means it is real symmetric positive semidefinite. 
We merely need to show it isn't the zero matrix since for real symmetric positive semidefinite $C$, we know $tr\big(C\big)\geq 0$ with equality iff $C=\mathbf 0$
(easy proof:  $tr\big(C\big) = tr\big(C^\frac{1}{2}C^\frac{1}{2}\big) = tr\big((C^\frac{1}{2})^TC^\frac{1}{2}\big) =\big\Vert C^\frac{1}{2}\big \Vert_F^2$ and the Frobenius norm is positive definite.  )  
I'd suggest finishing by noting that if we select any non-zero $\mathbf z$ where $\mathbf z^T \mathbf y =0$ then 
$\mathbf z^T B^{+}\mathbf z=\mathbf z^T B\mathbf z - \mathbf z^T\mathbf{yy}^T\mathbf z = \mathbf z^T B\mathbf z +0= \mathbf z^T B\mathbf z \gt 0$
so $B^{+} \neq \mathbf 0$ and hence has positive trace  
