For acute triangle with sides $a$, $b$, $c$ and circumradius $R$, define $x$, $y$, $z$ with $a+x=b+y=c+z=2R$. Show $x+y+z\lt2R$.

Question

This problem has been revealed to me not so easy as I would have thought.

Let $a,b,c$ be the sides of an acute triangle and $R$ its circumradius. It is clear that there are three positive numbers $x,y,z$ such that $a+x=b+y=c+z=2R$.

Prove that $x+y+z\lt2R$.