EDIT. 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. $abc\leq 0$. WORK IN PROGRESS.

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 !

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 !

The required result is true. 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 $\lambda_i,\mu_i> 0$. Here $D_1-K>0,D_2-K>0,\sqrt{D_1D_2}-K>0$ iff $\det(D_1-K)>0,\det(D_2-K)>0,\det(\sqrt{D_1D_2}-K)>0$ iff $f((\lambda_i))>0,f((\mu_i))>0,f((\sqrt{\lambda_i\mu_i}))>0$ where $f(u,v,w)=uvw-2abc-(vb^2+uc^2+wa^2)$ ; we must prove that the condition $f((\lambda_i))>0,f((\mu_i))>0$ implies that $f((\sqrt{\lambda_i\mu_i}))>0$.

Case 1. $\det(K)=2abc\leq 0$. Then $K\leq 0$ and we are done.

Case 2. $abc>0$. 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 $U,V,W,g(U,V,W),g(U',V',W')>0$ ; 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 $UV'+VU'\geq 2\sqrt{UU'VV'}$.

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 !

EDIT. 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. $abc\leq 0$. WORK IN PROGRESS.

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 !