$\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}=?$
In the figure, $I$ is the incenter of the triangle $ABC$ , $a$ and $b$ are maximal, and $r$ is the inradius of the triangle $MNI$.
I try
$a$(max) and $b$(max) implies a and b are at the midpoints of $AB$ and $BC$
$\triangle AJK_{(right)}:OM = c-a: ON = c-b $
$MN^2 = (c-a)^2+(c-b)^2 \implies MN^2 = c^2-2ca+a^2+c^2-2bc+b^2=2(c^2+a^2-ac-bc)$
$\triangle AFO: AF^2=c^2-(c-2a)^2\implies AF =2\sqrt{ac-a^2}$
$\triangle OHC: CH^2=c^2-(c-2b)^2\implies CH =2\sqrt{bc-b^2}$
I will use $\alpha, \beta,\gamma;\rho$ instead of the values $a,b,c;r$ from the statement above, since i have a better usage for the good old latin letters.
The solution below is analytic, i saw no striking geometric property of $\rho$ that would allow a simple computation for this length.
We will place the given triangle $\Delta ABC$ with two vertices $A=(4a,0)$ and $B=(0,4b)$ on the cartesian axes. Its right angle is in $C=(0,0)$. The factor four appears in order to minimize typing denominators. So the sides of $\Delta ABC$ have lengths $4a$, $4b$, and $4c$, connected by the usual pythagorean identity $$ c^2 = a^2+b^2\ . $$ We will use $a,b,c$ (and later $r$) with this meaning only.
Then $\alpha,\beta,\gamma$ have immediate formulas in terms of $a,b,c$: $$ \begin{aligned} \alpha &= c-b\ ,\\ \beta &= c-a\ ,\\ \gamma &= 2c\ . \end{aligned} $$ So we need to "relate somehow" the expression $$ \begin{aligned} E &= \frac 1{\sqrt{\alpha\beta}} + \frac 1{\sqrt{\beta\gamma}} + \frac 1{\sqrt{\gamma\alpha}} \\ &= \frac 1{\sqrt{(c-a)(c-b)}} + \frac 1{\sqrt{2c(c-a)}} + \frac 1{\sqrt{2c(c-b)}} \\ &= \frac {\sqrt{(c+a)(c+b)}}{ab} + \frac {\sqrt{2c(c+a)}}{2bc} + \frac {\sqrt{2c(c+b)}}{2ac} \\ &=\frac 1{2abc}\Big(\ 2c\sqrt{(c+a)(c+b)} + a\sqrt{2c(c+a)} + b\sqrt{2c(c+b)} \ \Big) \\ &=\frac {\sqrt2}{2abc}\Big(\ c(a+b+c) + a\sqrt{c(c+a)} + b\sqrt{c(c+b)} \ \Big) \ . \end{aligned} $$ to the length $\rho$, best solely as a function of $\rho$.
We will compute the distance $IJ$
The inradius $r$ is computed as area $\frac 12 4a\cdot 4b$ divided by half-perimeter $\frac12 (4a+4b+4c)$. This gives: $$ r=\frac {4ab}{a+b+c}\ .$$ The cartesian equation of the line $MN$ w.r.t. the axes $CA=Ox$ and $CB=Oy$ is $$ 0= \begin{vmatrix} 1 & x & y\\ 1 & 2a & b-c\\ 1 & a-c & 2b \end{vmatrix} \ . $$ Expanding and norming to an equation of the shape $ux+vy+w=0$ with $u^2+v^2=1$ we obtain $$ \begin{aligned} u &=(b+c)/q\ ,\\ v &=(a+c)/q\ ,\\ w &=-\det\begin{bmatrix} 2a & b-c\\ a-c & 2b \end{bmatrix} /q %= (4ab-(c-a)(c-b))/q \ ,\\ &\qquad\text{ where}\\ q&=\sqrt{(a+c)^2+(b+c)^2}\ . \end{aligned} $$ The distance from $I$ to $MN$ is obtained by plugging in $I=(r,r)$ as $(x,y)$ inside $|ux+vy+w|$, so the height $IJ$, and the sides of $\Delta IMN$ are determined by: $$ \begin{aligned} IJ &=\frac1q \Big(\ r(a+b+2c)-(3ab+ac+bc-c^2)\ \Big)\ ,\\ MN^2 &= (2c-\beta)^2 + (2c-\alpha)^2 = (a+c)^2+(b+c)^2 = q^2\ ,\\ MI^2 &= (r-2a)^2 + (r+\alpha)^2\ ,\\ NI^2 &= (r-2b)^2 + (r+\beta)^2\ .\\ &\qquad\text{ Let make these expressions explicit:}\\ MN &= q\\ &=\sqrt{c(2a+2b+3c)} \ ,\\[3mm] MN\cdot IJ\;(a+b+c) &= 4ab(a+b+2c) - (a+b+c)(3ab+ac+bc-c^2) \\ &=(4ab-3ab)(a+b+c) +4abc -(a+b+c)c(a+b-c)\\ &=ab(a+b+c)+4abc - c((a+b)^2-c^2)\\ &=ab(a+b+c)+4abc - c(2ab)\\ &=ab(a+b+3c)\ ,\\ MI^2(a+b+c)^2 &= (4ab- 2a(a+b+c))^2 + (4ab +(c-b)(a+b+c))^2 \\ &=4a^2(a-b+c)^2 + (4ab+a(c-b)+(c^2-b^2))^2 \\ &=4a^2(a-b+c)^2 + (3ab + ac + a^2)^2 \\ &=4a^2(a-b+c)^2 + a^2(a+3b+c)^2\\ &=a^2(5a^2 - 2ab + 13b^2 + 10ac - 2bc + 5c^2)\qquad\text{ replace }b^2=c^2-a^2\\ &=a^2(-8a^2 - 2ab + 10ac - 2bc + 18c^2)\\ &=2a^2(-8a^2 - 2ab + 10ac - 2bc + 18c^2)\\ &=2a^2(a+c)(9c-4a-b) \ , \\[3mm] NI^2(a+b+c)^2 &= 2b^2(b+c)(9c-a-4b) \ . \\[3mm] % &\qquad\text{ From here we compute:}\\ \rho &= \frac {\operatorname{Area}(\Delta MNI)} {\operatorname{Halfperimeter}(\Delta MNI)} =\frac{MN\cdot IJ/2}{(MN+MI+NI)/2} \\ \frac 1\rho &= \frac {(MN+MI+NI)(a+b+c)}{MN\cdot IJ\; (a+b+c)} \\ &= \frac 1{ab(a+b+3c)} \Bigg( (a+b+c)\sqrt{c(2a+2b+3c)} \\ &\qquad\qquad\qquad\qquad\qquad\qquad + a\sqrt{2(a+c)(9c-4a-b)} \\ &\qquad\qquad\qquad\qquad\qquad\qquad + b\sqrt{2(b+c)(9c-a-4b)} \Bigg)\ . \end{aligned} $$ And now i cannot see how to express $E$ only in terms of $\rho$.
Check:
A picture check first, we are using the special values $a=2$, $b=3$, $c=\sqrt{13}$, and the computed value for $\rho$ is:
Computer check for the numerical values in the picture, using two formulas for $\rho$, the simpler one from the start, and the more complicated one obtained after reshaping algebraically to get a final form.
And the numerical values are:
Let us consider some other special situation, $a=9$, $b=40$, $c=41$. We compute $E$ and $\rho$ in this case:
The expression $E$ is:
So in this special case $$ E = \frac 1{656}(2\sqrt{41} + 8\sqrt{82} + 82\sqrt 2)\ . $$
The value for $\rho$ is
So $$ \frac 1\rho = \frac 1{1376}(2\sqrt{293} +2\sqrt{9061} +160)\ . $$ An algebraic formula for $E$ (in this special case) involving only $\rho$ is excluded, since we are working in a different number field.