Why does $\left\| {\left| A \right|} \right\| \le \left\| {\left| I \right|} \right\|$, for every doubly stochastic matrix $A \in M_n$? Let $\left\| {\left| . \right|} \right\|$ be a unitarily invariant matrix norm on $M_n$.
Why does $\left\| {\left| A \right|} \right\| \le \left\| {\left| I \right|} \right\|$, for every doubly stochastic matrix $A \in M_n$ ?

In the spirit of Omnomnomnom's answer, here is an alternative approach that doesn't require Birkhoff theorem.
It is well-known that when $A$ is doubly stochastic, its operator norm is equal to $1$ (because $1\le\rho(A)^2 \le \|A\|_2^2 = \rho(A^TA) \le \|A^TA\|_1 = 1$). So, by unitary invariance of the norm $|||\cdot|||$ in question and by singular value decomposition, you may assume that $A$ is a nonnegative diagonal matrix whose largest diagonal entry is $1$. It is easy to show that such a matrix is a convex combination of at most $2n$ diagonal real orthogonal matrices.
