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I am new here and this my first question, I hope I am being as clear as possible and apologize in advance for any misunderstandings.

I am researching the Arithmetic-Geometric Mean (AM-GM) inequality for matrices. Which can be seen as an "extension" of the obvious AM-GM inequality for positive numbers $a,b$ : $ \sqrt{ab} \leq \frac{1}{2}(a+b) $. One of such possible extensions is the following:

$|||A^{1/2}B^{1/2}|||\leq \frac{1}{2}|||A+B|||$ As shown in "Positive Definite Matrices" by R.Bhatia.

Where A, B are positive, hermitian matrices, and $||| \cdot |||$ is a unitarily invariant norm.

This is precisely my problem/question: all proofs and attempts to investigate the AM-GM matrix-norm-inequality use unitarily invariant norms. Does this mean that for non-unitarily invariant norms this inequality does not hold? Is there a counterexample for this? perhaps for p-norms such as $\|\cdot\|_\infty$ , $\|\cdot\|_1$? Or an explanation why a proof for non-unitarily invariant norms could not work?

I am just lost since all the texts I read regarding the AM-GM inequality, immediately jump to using unitarily invariant norms.

Thank you!!

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Consider $$A = \begin{pmatrix} 46 & 7 & 8 \\ 7 & 36 & 16 \\ 8 & 16 & 30\end{pmatrix}\qquad \text{ and }\qquad B=\begin{pmatrix} 34 & 7 & 12 \\ 7 & 48 & 10 \\ 12 & 10 & 44\end{pmatrix}.$$ Then $$\frac{1}{2}\|A+B\|_1 =\frac{1}{2}\|A+B\|_{\infty}=62$$ however $$ \|A^{1/2}B^{1/2}\|_1\approx 62.1527 \qquad \text{ and }\qquad \|A^{1/2}B^{1/2}\|_{\infty}\approx 62.1037.$$

Here is the Matlab code to generate such counter examples:

n = 3;  % choose the dimension of the matrices
stop = 0;  
while stop == 0  
  % Generates a pair of symmetric positive definite matrices
  A = floor(10*rand(n));  
  A=A+A';  
  A = A+10*n*eye(n);  
  B = floor(10*rand(n));  
  B = B+B';  
  B = B+10*n*eye(n);  
  % Computes their square root
  AS = A^(1/2);  
  BS=B^(1/2);
  % Check if the AM-GM is not satisfied
  if norm(AS*BS,1)>norm(A+B,1)/2 && norm(AS*BS,inf)>norm(A+B,inf)/2  
    stop =1;  
  end  
end
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