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I asked the following question here: "Does there exist $p>1$ such that for all $n\geq 2$, if $(a_{ij})$ and $(b_{ij})$ are symmetric positive semidefinite $n\times n$ matrices and $a_{ij}, b_{ij}\geq 0$ then $(\|(a_{ij},b_{ij})\|_p)=((a_{ij}^p+b_{ij}^p)^{1/p})$ is also positive semidefinite? Maybe, a simpler question: is it true for $p=2$?"

Paata Ivanishvili gave a counterexample for $n=3$ for all $p>1$. However, under an additional constraint that all $a_{ii}=b_{ii}=1$, he showed that the answer is positive for $p=2, n=3$. So, my follow-up question is: under the constraints

$a_{ij}, b_{ij}\geq 0, a_{ii}=b_{ii}=1, p=2,$

is the above matrix positive semidefinite for all $n\geq 2$?

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    $\begingroup$ This is getting boring. Consider the quadratic forms $(x_1+x_2)^2+(x_3+x_4)^2$ and $(x_1+x_3)^2+(x_2+x_4)^2$ and the vector $(1,-1,-1,1)$. $\endgroup$
    – fedja
    Jul 15, 2019 at 1:42

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