Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\|\cdot\|)$.
For $A \equiv V|A|$ the canonical polar decomposition of $A\in \mathcal{B}(F)$, we let $\tilde{A}:=|A|^{1/2}V|A|^{1/2}$. Here $|A|=\sqrt{A^*A}$ and $V$ is a partial isometry.
I want to prove that $$\|\tilde{A}\|\leq \|A\|.$$
We have $$\|\tilde{A}\|=\sup\{\||A|^{1/2}V|A|^{1/2}x\|: x\in F, \|x\|=1 \}$$
We have $\|A^* A\| = \|A\|^2$ (the $C^* $ identity), and $|A|$ is the positive semidefinite square root of $A^* A$ so (using the spectral theorem) $\||A|\| = \|A\|$ and $\||A|^{1/2}\| = \|A\|^{1/2}$. Then $$ \|\tilde{A}\| \le \||A|^{1/2}\| \|V\| \||A|^{1/2}\| = \|A\|$$