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- By the Lie product formula, .
By the Golden inequality, . (en)
- For any , we have , thus .
Thus is a contraction map, thus , thus , thus all eigenvalues of are nonpositive, thus . (en)
- If for all , then all the other inequalities are also proven as special cases of it. So it suffices to prove that inequality.
case is trivial.
case. Since are Hermitian and PSD, we can split to , which allows us to write , meaning it is a non-negative real number.
Now by Cauchy–Schwarz inequality,
case. Define two sequences of matrices which, by construction, are Hermitian and positive semidefinite.
For any , by the cyclic property of trace,
By the same argument as case, . Apply Cauchy–Schwarz, and the cyclic equalities,
If , then .
Otherwise, by induction, and continuing the same argument,. This continues until we obtain . (en)
- It suffices to show that .
by the Golden inequality.
The second claim is proven similarly. (en)
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