#Python 3, 66 bytes

This formerly took advantage of the new assignment expressions ("walrus operator") introduced in Python 3.8. Thanks to commentors, I've taken that out, so it works on previous versions too!

lambda a,b,c:((a*b*c/(b+c-a)/(a+c-b)/(a+b-c)-1)*a*b*c/(a+b+c))**.5

Try it online!

It's based on the same calculations described in Arnauld's answer, but using the perimeter \$p\$ instead of the semiperimeter \$s\$: $$ \begin{aligned}\\ p&=a+b+c\\ &=2s \end{aligned}\\ \therefore d=\sqrt{R^2-\frac{abc}{p}}\\ \text{and } R^2=\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)} $$

The grand total savings of this rearrangement is... two bytes.

Factoring \$p\$ out and expanding the terms in the denominator means I don't have to store \$p\$, saving another three bytes. I also stored the product \$abc\$ in a variable \$m\$, which saved some bytes at first... but it could later be factored out, turning the brackets-and-walrus into a liability, not a savings! Here's the final formula: \begin{aligned} d&=\sqrt{\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)}-\frac{abc}{p}}\\ &=\sqrt{\left(\frac{abc}{(b+c-a)(a+c-b)(a+b-c)}-1\right)\frac{abc}{p}} \end{aligned}

Python 3, 66 bytes

This formerly took advantage of the new assignment expressions ("walrus operator") introduced in Python 3.8. Thanks to commentors, I've taken that out, so it works on previous versions too!

lambda a,b,c:((a*b*c/(b+c-a)/(a+c-b)/(a+b-c)-1)*a*b*c/(a+b+c))**.5

Try it online!

It's based on the same calculations described in Arnauld's answer, but using the perimeter \$p\$ instead of the semiperimeter \$s\$: $$ \begin{aligned}\\ p&=a+b+c\\ &=2s \end{aligned}\\ \therefore d=\sqrt{R^2-\frac{abc}{p}}\\ \text{and } R^2=\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)} $$

The grand total savings of this rearrangement is... two bytes.

Factoring \$p\$ out and expanding the terms in the denominator means I don't have to store \$p\$, saving another three bytes. I also stored the product \$abc\$ in a variable \$m\$, which saved some bytes at first... but it could later be factored out, turning the brackets-and-walrus into a liability, not a savings! Here's the final formula: \begin{aligned} d&=\sqrt{\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)}-\frac{abc}{p}}\\ &=\sqrt{\left(\frac{abc}{(b+c-a)(a+c-b)(a+b-c)}-1\right)\frac{abc}{p}} \end{aligned}

#Python 3, 67 bytes

This formerly took advantage of the new assignment expressions ("walrus operator") introduced in Python 3.8. Thanks to commentors, I've taken that out, so it works on previous versions too!

lambda a,b,c:((a*b*c/(b+c-a)/(a+c-b)/(a+b-c)-1)*a*b*c/(a+b+c))**.5

Try it online!

It's based on the same calculations described in Arnauld's answer, but using the perimeter \$p\$ instead of the semiperimeter \$s\$: $$ \begin{aligned}\\ p&=a+b+c\\ &=2s \end{aligned}\\ \therefore d=\sqrt{R^2-\frac{abc}{p}}\\ \text{and } R^2=\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)} $$

The grand total savings of this rearrangement is... two bytes.

Factoring \$p\$ out and expanding the terms in the denominator means I don't have to store \$p\$, saving another three bytes. I also stored the product \$abc\$ in a variable \$m\$, which saved some bytes at first... but it could later be factored out, turning the brackets-and-walrus into a liability, not a savings! Here's the final formula: \begin{aligned} d&=\sqrt{\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)}-\frac{abc}{p}}\\ &=\sqrt{\left(\frac{abc}{(b+c-a)(a+c-b)(a+b-c)}-1\right)\frac{abc}{p}} \end{aligned}

#Python 3, 66 bytes

This formerly took advantage of the new assignment expressions ("walrus operator") introduced in Python 3.8. Thanks to commentors, I've taken that out, so it works on previous versions too!

lambda a,b,c:((a*b*c/(b+c-a)/(a+c-b)/(a+b-c)-1)*a*b*c/(a+b+c))**.5

Try it online!

It's based on the same calculations described in Arnauld's answer, but using the perimeter \$p\$ instead of the semiperimeter \$s\$: $$ \begin{aligned}\\ p&=a+b+c\\ &=2s \end{aligned}\\ \therefore d=\sqrt{R^2-\frac{abc}{p}}\\ \text{and } R^2=\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)} $$

The grand total savings of this rearrangement is... two bytes.

Factoring \$p\$ out and expanding the terms in the denominator means I don't have to store \$p\$, saving another three bytes. I also stored the product \$abc\$ in a variable \$m\$, which saved some bytes at first... but it could later be factored out, turning the brackets-and-walrus into a liability, not a savings! Here's the final formula: \begin{aligned} d&=\sqrt{\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)}-\frac{abc}{p}}\\ &=\sqrt{\left(\frac{abc}{(b+c-a)(a+c-b)(a+b-c)}-1\right)\frac{abc}{p}} \end{aligned}

#Python 3.8, 67 bytes

This takes advantage of the new assignment expressions ("walrus operator") introduced in Python 3.8.

lambda a,b,c:(((m:=a*b*c)*m/(b+c-a)/(a+c-b)/(a+b-c)-m)/(a+b+c))**.5

Try it online!

It's based on the same calculations described in Arnauld's answer, but using the perimeter \$p\$ instead of the semiperimeter \$s\$: $$ \begin{aligned}\\ p&=a+b+c\\ &=2s \end{aligned}\\ \therefore d=\sqrt{R^2-\frac{abc}{p}}\\ \text{and } R^2=\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)} $$

The grand total savings of this rearrangement is... two bytes. Storing the product \$abc\$ in \$m\$ saves another one two bytes.

Factoring \$p\$ out and expanding the terms in the denominator means I don't have to store \$p\$, saving another three bytes: \begin{aligned} d&=\sqrt{\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)}-\frac{abc}{p}}\\ &=\sqrt{\left(\frac{\left(abc\right)^2}{(b+c-a)(a+c-b)(a+b-c)}-abc\right)\frac{1}{p}} \end{aligned}

#Python 3, 67 bytes

This formerly took advantage of the new assignment expressions ("walrus operator") introduced in Python 3.8. Thanks to commentors, I've taken that out, so it works on previous versions too!

lambda a,b,c:((a*b*c/(b+c-a)/(a+c-b)/(a+b-c)-1)*a*b*c/(a+b+c))**.5

Try it online!

It's based on the same calculations described in Arnauld's answer, but using the perimeter \$p\$ instead of the semiperimeter \$s\$: $$ \begin{aligned}\\ p&=a+b+c\\ &=2s \end{aligned}\\ \therefore d=\sqrt{R^2-\frac{abc}{p}}\\ \text{and } R^2=\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)} $$

The grand total savings of this rearrangement is... two bytes.

Factoring \$p\$ out and expanding the terms in the denominator means I don't have to store \$p\$, saving another three bytes. I also stored the product \$abc\$ in a variable \$m\$, which saved some bytes at first... but it could later be factored out, turning the brackets-and-walrus into a liability, not a savings! Here's the final formula: \begin{aligned} d&=\sqrt{\frac{\left(abc\right)^2}{p(p-2a)(p-2b)(p-2c)}-\frac{abc}{p}}\\ &=\sqrt{\left(\frac{abc}{(b+c-a)(a+c-b)(a+b-c)}-1\right)\frac{abc}{p}} \end{aligned}