 Solution

Submultiplicativity and the C-star identity give $\Vert x{\Vert }^2=\Vert x^{\ast }x\Vert \le \Vert x^{\ast }\Vert \Vert x\Vert$. For $x\ne 0$ this implies $\Vert x\Vert \le \Vert x^{\ast }\Vert$. Apply the same argument to $x^{\ast }$ and use $x^{\ast \ast }=x$ for the reverse inequality; $x=0$ is immediate. Hence