= Solution
The final printed direction is reversed under the <majorization> convention stated in the same part. A completely depolarizing <unital quantum channel> sends a pure qubit state to $I/2$, giving spectra $r=(1,0)$ and $s=(1/2,1/2)$. The proposed $r\prec s$ fails already at the first partial sum, since $1>1/2$. The valid conclusion is
$$
\boxed{s\prec r.}
$$
Here is its general proof. Choose eigenbases $\rho=\sum_i r_i|u_i\rangle\langle u_i|$ and $\sigma=\sum_j s_j|v_j\rangle\langle v_j|$, and define the <eigenvalue mixing matrix of a unital quantum channel>
$$
A_{ji}=\langle v_j|\Lambda(|u_i\rangle\langle u_i|)|v_j\rangle.
$$
Its entries are nonnegative by positivity. Trace preservation gives $\sum_jA_{ji}=1$ for every column. Unitality gives $\sum_iA_{ji}=\langle v_j|\Lambda(I)|v_j\rangle=1$ for every row. Thus $A$ is a <doubly stochastic matrix>, and linearity gives
$$
s_j=\sum_iA_{ji}r_i,\qquad s=Ar.
$$
The supplied doubly stochastic characterization of <majorization> then yields $s\prec r$, completing the intended corrected result. This proves <spectral majorization under a unital quantum channel>. A <unital quantum channel> can make a spectrum more uniform; it cannot generally make the output spectrum majorize the input. The counterexample resolves the false printed assertion rather than treating the reversed inequality as a convention change.
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