Is $x\mapsto \sqrt{xx^*}$ a norm? A real $*$-algebra $X$ (real algebra with linear operation $*$ such that $(xy)^* = y^*x^*$ and $x^{**} = x$ for all $x,y\in X$) is said to be nicely normed if $X$ is unital and the operation $*$ has the following additional properties:
$$xx^* = x^*x \geq 0 \quad \text{and} \quad x+x^* \in \mathbb{R}$$
for all $x\in X$ with $xx^* = 0 \Leftrightarrow x = 0$.
Define $||x|| = \sqrt{xx^*}$. I cannot prove that this is a norm; my problem is only the triangular inequality. I have tried with $\text{Re} (x) = (x+x^*)/2$ and $\text{Im} (x) = (x-x^*)/2$ (it can be seen that $||x||^2 = \text{Re}^2(x)- \text{Im}^2(x)$)
and with the direct approach, but I did not succeed.
 A: After the comments to my earlier reply (which was not satisfactory for the OP's purposes), I emailed John Baez to ask for a proof. He very kindly responded as follows (paraphrasing ever so slightly): 
"The definition of 'nicely normed' includes the fact that $aa^* > 0$ for any nonzero $a$.  This means $aa^*$ can be identified with a positive real number.  Doing this, it follows that the mapping $a\mapsto aa^*$ is a positive definite quadratic form.  It follows that the square root of this obeys the triangle inequality.   (Up to an invertible linear transformation, any positive definite quadratic form on a finite-dimensional vector space is the square of the usual Euclidean norm on $\mathbb{R}^n$.)"
(Many thanks to John for this answer.) 
A: Here is a proof. First, we have $\|tx\| = |t|\cdot\|x\|$ for scalars $t$ (easy).  
In this case, the triangle inequality is equivalent to the fact that the unit ball $\{v \in X: \|v\| \leq 1\}$ is convex, viz., if $\|v\|, \|w\| \leq 1$, then $\|tv + (1-t)w\| \leq 1$ when $0 \leq t \leq 1$. In one direction, to prove that the inequality $\|x + y\| \leq \|x\| + \|y\|$ follows from convexity, suppose without loss of generality that $x, y \neq 0$, and put $v = x/\|x\|$, $w = y/\|y\|$, and $t = \|x\|/(\|x\|+\|y\|)$, observing that $(x+y)/(\|x\|+\|y\|) = tv+(1-t)w$ has norm less than $1$ by convexity. 
To prove the unit ball is convex, start from $\|x\|^2 = \text{Re}(x)^2 - \text{Im}(x)^2$, borrowing notation from the OP. Note that 


*

*$(2\text{Im}(x))^2 = (x-x^\ast)^2 \leq 0$. Proof: if $y = x-x^\ast$, then $y^\ast = -y$, so $-y^2 = y^\ast y \geq 0$. 


It follows that the real scalar $\text{Re}(x)$ satisfies $|\text{Re}(x)| \leq \|x\|$. 
Next, we have $\|x y\|^2 = x y y^\ast x^\ast = (y y^\ast)(x x^\ast)$ (since $y y^\ast$ is a real scalar, it commutes with any $x$), so $\|xy\| = \|x\|\cdot \|y\|$.
Putting all this together, we show convexity of the unit ball. Suppose $x, y$ have norm $1$. Then for $t \in [0, 1]$, we have 
$$\|tx + (1-t)y\|^2 = (tx + (1-t)y)(tx^\ast + (1-t)y^\ast) = t^2\|x\|^2 + t(1-t)(x y^\ast + y x^\ast) + (1-t)^2\|y\|^2 \qquad (1)$$ 
where $x y^\ast + y x^\ast = 2\text{Re}(x y^\ast) \leq 2\|x y^\ast\| = 2\|x\|\cdot \|y\| = 2$. Applying this inequality to equation (1), we get 
$$\|tx + (1-t)y\|^2 \leq t^2 + 2t(1-t) + (1-t)^2 = 1$$ 
which completes the proof. 
