About triangle inequality in C*algebras In Euclidean spaces we can say triangle inequality 
$\| {a+b}\|\leq \|{a}\|+\|{b}\|$
becomes equality $\|a+b\|=\|a\|+\|b\|$
When $ a=t b $ for some positive t.
What about C$^*$-algebras? We know that the set of positive elements is a cone and the unit ball is the closed convex hull of the unitaries. 
Now if $a$ and $b$ are two positive elements in a C$^*$-algebra such that $\|a\|=\|b\|=1$, and $ \|a+b\|=\|a\|+\|b\|.$ what can I tell about $a$ and $b$ here?
 A: In a $C^*$-algebra if $a = tb$ for some $t>0$ you still have that $\|a + b\| = \|tb + b\| = |t+1|\cdot\|b\|=|t|\cdot\|b\|+\|b\| = \|a\| + \|b\|$. 
Going the other way there is not much you can say. Even in the commutative case if we have $f,g\in C(X)$ such that $argmax(f) = argmax(g) = x_0\in X$. Then $\|f+g\| = f(x_0)+g(x_0) = \|f\| + \|g\|$. For unital $C^*$-algebras (ie compact $X$) you can probably show that this is an if and only if (ie $\|f + g\| = \|f\| + \|g\|$ if and only if $f$ and $g$ achieve their max at the same points, but beyond that I don't think you can say anything. 
A: Thinking of the elements of the C$^*$-algebra as operators on a Hilbert space, we can reason as follows. 
Assume first that $\|a+b\|$ is an  eigenvalue of $a+b$. Then, if $x$ is a unit eigenvector, 
$$
\|a+b\|=\langle (a+b)x,x\rangle=\langle ax,x\rangle+\langle bx,x\rangle\leq\|a\|+\|b\|.
$$
The equality forces $\|a\|=\langle ax,x\rangle$ and $\|b\|=\langle bx,x\rangle$, so $\|a\|$ and $\|b\|$ are eigenvalues of $a$ and $b$, respectively, with the same eigenvector $x$. In general, an approximate version of this is possible. 
Beyond that, not much more can be said. The operator norm is a supremum norm, and as such it forgets lots of parts of the operator. For instance, let $C,D$ be arbitrary positive operators with $0\leq C,D\leq I$. Then $a=1\oplus C$ and $b=1\oplus D$ satisfy $\|a\|=\|b\|=1$, $\|a+b\|=2$. 
