Suppose $D\_1$ and $D\_2$ are two $3\times 3$ diagonal matrices with real positive entries on their diagonal. Let $K$ be a symmetric $3\times 3$ matrix with zeroes on its diagonal but with arbitrary real entries elsewhere. If $D\_1 > K$ and $D\_2 > K$ then is $\sqrt{D\_1 D\_2} > K ?$ (where $A>B$ if $A-B$ is positive-definite).



Geometric mean of two matrices
EDIT 2. The required result is FALSE. cf. Case 2 below. 


Let $D\_1=diag((\lambda\_i)),D\_2=diag((\mu\_i)),K=\begin{pmatrix}0&a&b\\a&0&c\\b&c&0\end{pmatrix}$ where $\lambda\_i,\mu\_i> 0$. Assume that $D\_1-K>0,D\_2-K>0$ ; we want to show $\sqrt{D\_1D\_2}-K>0$. Clearly $\lambda\_1\lambda\_2>a^2,\mu\_1\mu\_2>a^2$ implies $\sqrt{\lambda\_1\mu\_1}\sqrt{\lambda\_2\mu\_2}>a^2$. 


Thus it remains to show that $\det(D\_1-K)>0,\det(D\_2-K)>0$ implies that $\det(\sqrt{D\_1D\_2}-K)>0$ ; that is $f((\lambda\_i))>0,f((\mu\_i))>0$ implies that $f((\sqrt{\lambda\_i\mu\_i}))>0$ where $f(u,v,w)=uvw-2abc-(vb^2+uc^2+wa^2)$. 


Case 1. $abc>0$.
If $\lambda b^2=L,\mu b^2=M$, then $\sqrt{\lambda\mu}b^2=LM$ ; thus we may put $vb^2=Vabc,\cdots$, that is $v=\dfrac{ac}{b}V,\cdots$. We obtain $g(u,v,w)=\dfrac{f(u,v,w)}{abc}=UVW-(U+V+W)-2$. Assume that $U,V,W,U',V',W',g(U,V,W),g(U',V',W')>0$ ; we must show that $g(\sqrt{UU'},\sqrt{VV'},\sqrt{WW'})>0$. Since $UU'VV'WW'>(2+U+V+W)(2+U'+V'+W')$, it remains to show that $(2+U+V+W)(2+U'+V'+W')\geq (2+\sqrt{UU'}+\sqrt{VV'}+\sqrt{WW'})^2$ ; it is true because $U+U'\geq 2\sqrt{UU'}$ and $UV'+VU'\geq 2\sqrt{UU'VV'}$.


Case 2. $abc\leq 0$. That follows is a counter-example. Let $K=\begin{pmatrix}0&-1&1\\-1&0&1\\1&1&0\end{pmatrix}$, $D\_1=diag(682.0000874343,0.0615388680257,16.5992335482)$, $D\_2=diag(53.1443607121,0.5046212602,2.0005213111)$. One has $D\_1-K>0,D\_2-K>0$ ; yet $\sqrt{D\_1D\_2}-K$ has an eigenvalue $\approx -9.10^{-4}$ and we are done !


REMARK . Note that $D\_1D\_2-K^2>0$ is false (in general). For instance, take $D\_1=D\_2=diag(1,2,3),K=\begin{pmatrix}0&1&1\\1&0&0\\1&0&0\end{pmatrix}$ ; then $D\_1=D\_2>K$ and $D\_1D\_2-K^2$ is not positive. More generally, let $U\geq V\geq 0$ ; then $U^2\geq V^2$ is not necessarily true ; yet $\sqrt{U}\geq \sqrt{V}$ is always true !


