Let $x,y,z$ be elements of a C*-algebra $A$. Does the inequality $$|xyz| \leq |xz|^\frac{1}{2} | y| \ | xz|^\frac{1}{2}$$ hold in the positive cone of $A$? Here $|x| := (x^*x)^\frac{1}{2}$.
If $y= 1$, this is an equality.
If $z=1$, this is $ |xy| \leq |x|^\frac{1}{2} |y| |x|^\frac{1}{2}$, which I'm already not too sure about.
Hmm it seems the answer must be "no".
Take $z = 1$, and $x$ to be a projection $e$. So, the inequality to be proved becomes $$|ey| \leq e | y| e.$$
Now, take $y$ to be a partial isometry $w$ with $ww^* = e$ and $w^*w = e'$ where $e'$ is orthogonal to $e$. Then, $|ey| = e$ while $e|y| e = ee'e = 0 $, but $e \leq 0$ fails (assuming $e \neq 0$).
Sorry for the apparently silly question...
Although it's probably quite clear, I guess I'd may as well add that the counterexample outlined above already crops up when $A = \operatorname{Mat}_{2 \times 2}(\mathbb{C})$. Just take \begin{align*} x = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} && y = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} && z = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{align*} to see the inequality fail.