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This is merely a variation of your own proof, Matt, but I believe it makes things clearer.

The first step is to define $c:=1-a-b$. Then, your identity takes the form

$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}=\pi^{1/2}\dfrac{\Gamma\left(\dfrac{a}{2}\right)}{\Gamma\left(\dfrac{1-a}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{b}{2}\right)}{\Gamma\left(\dfrac{1-b}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{c}{2}\right)}{\Gamma\left(\dfrac{1-c}{2}\right)}$

for $a+b+c=1$. This is symmetric in $a$, $b$, $c$, which means we are way less likely to go insane during the following computations.

Now, using the formula

$\dfrac{\Gamma\left(\dfrac{s}{2}\right)}{\Gamma\left(\dfrac{1-s}{2}\right)}=\pi^{-1/2}2^{1-s}\cdot\cos\dfrac{\pi s}{2}\cdot\Gamma\left(s\right)$,

the right hand side simplifies to

$\pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$

(here we used $2^{3-a-b-c}=2^{3-1}=4$), so the identity in question becomes

$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}$ $= \pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$.

Dividing by $\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(c\right)$ on both sides, we get

$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(b+c\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(c+a\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(a+b\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.

Since $b+c=1-a$, $c+a=1-b$, $a+b=1-c$, this rewrites as

$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(1-a\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(1-b\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(1-c\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.

Now, using the formula $\dfrac{1}{\Gamma\left(z\right)\Gamma\left(1-z\right)}=\pi^{-1}\sin{\pi z}$ on the left hand side, and dividing by $\pi^{-1}$, we can simplify this to

$\sin{\pi a}+\sin{\pi b}+\sin{\pi c}=4\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.

Since $a+b+c=1$, we can set $A=\pi a$, $B=\pi b$, $C=\pi c$ and then have $A+B+C=\pi$. Our goal is to show that

$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$

for any three angles $A$, $B$, $C$ satisfying $A+B+C=\pi$.

Now this can be proven in different ways:

1) One is by writing $C=\pi-A-B$ and simplifying using trigonometric formulae; this is rather boring and it breaks the symmetry.

2) Another one is using complex numbers: let $\alpha=e^{iA/2}$, $\beta=e^{iB/2}$ and $\gamma=e^{iC/2}$. Then,

$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$

becomes

(1) $\dfrac{\alpha^2-\alpha^{-2}}{2i}+\dfrac{\beta^2-\beta^{-2}}{2i}+\dfrac{\gamma^2-\gamma^{-2}}{2i} = 4\cdot\frac{\alpha+\alpha^{-1}}{2}\cdot\frac{\beta+\beta^{-1}}{2}\cdot\frac{\gamma+\gamma^{-1}}{2}$4\cdot\dfrac{\alpha+\alpha^{-1}}{2}\cdot\dfrac{\beta+\beta^{-1}}{2}\cdot\dfrac{\gamma+\gamma^{-1}}{2}$. Oh, and$A+B+C=\pi$becomes$\alpha\beta\gamma=2i$. Now proving (1) is just a matter of multiplying out the right hand side and looking at the$8$terms (two of them, namely$\alpha\beta\gamma$and$\alpha^{-1}\beta^{-1}\gamma^{-1}$, cancel out, being$i$and$-i$, respectively). 3) Here is how I would have done it 8 years ago: We can WLOG assume that$A$,$B$,$C$are the angles of a triangle (this means that$A$,$B$,$C$lie in the interval$\left[0,\pi\right]$, additionally to satisfying$A+B+C=\pi$), because everything is analytic (or by casebash). We denote the sides of this triangle by$a$,$b$,$c$(so we forget about the old$a$,$b$,$c$), its semiperimeter$\dfrac{a+b+c}{2}$by$s$, its area by$\Delta$and its circumradius by$R$. Then,$\sin A=\dfrac{a}{2R}$(by the Extended Law of Sines) and similarly$\sin B=\dfrac{b}{2R}$and$\sin C=\dfrac{c}{2R}$, so that$\sin A+\sin B+\sin C=\dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}=\frac{a+b+c}{2R}=\frac{s}{R}$C=\dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}=\dfrac{a+b+c}{2R}=\dfrac{s}{R}$. On the other hand, one of the half-angle formulas shows that $\cos\dfrac{A}2=\sqrt{\dfrac{s\left(s-a\right)}{bc}}$, and similar formulas hold for $\cos\dfrac{B}2$ and $\cos\dfrac{C}2$, so that

$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$

$=4\cdot\sqrt{\dfrac{s\left(s-a\right)}{bc}}\cdot\sqrt{\dfrac{s\left(s-b\right)}{ca}}\cdot\sqrt{\dfrac{s\left(s-c\right)}{ab}}$

$=\frac{4s}{abc}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.=\dfrac{4s}{abc}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. Now,$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\Delta$(by Heron's formula) and$\Delta=\frac{abc}{4R}$\Delta=\dfrac{abc}{4R}$ (by another formula for the area of the triangle), so tis becomes

$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}=\dfrac{4s}{abc}\cdot\dfrac{abc}{4R}=\dfrac{s}{R}$.

This is exactly what we got for the left hand side, qed.

1

This is merely a variation of your own proof, Matt, but I believe it makes things clearer.

The first step is to define $c:=1-a-b$. Then, your identity takes the form

$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}=\pi^{1/2}\dfrac{\Gamma\left(\dfrac{a}{2}\right)}{\Gamma\left(\dfrac{1-a}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{b}{2}\right)}{\Gamma\left(\dfrac{1-b}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{c}{2}\right)}{\Gamma\left(\dfrac{1-c}{2}\right)}$

for $a+b+c=1$. This is symmetric in $a$, $b$, $c$, which means we are way less likely to go insane during the following computations.

Now, using the formula

$\dfrac{\Gamma\left(\dfrac{s}{2}\right)}{\Gamma\left(\dfrac{1-s}{2}\right)}=\pi^{-1/2}2^{1-s}\cdot\cos\dfrac{\pi s}{2}\cdot\Gamma\left(s\right)$,

the right hand side simplifies to

$\pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$

(here we used $2^{3-a-b-c}=2^{3-1}=4$), so the identity in question becomes

$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}$ $= \pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$.

Dividing by $\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(c\right)$ on both sides, we get

$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(b+c\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(c+a\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(a+b\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.

Since $b+c=1-a$, $c+a=1-b$, $a+b=1-c$, this rewrites as

$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(1-a\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(1-b\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(1-c\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.

Now, using the formula $\dfrac{1}{\Gamma\left(z\right)\Gamma\left(1-z\right)}=\pi^{-1}\sin{\pi z}$ on the left hand side, and dividing by $\pi^{-1}$, we can simplify this to

$\sin{\pi a}+\sin{\pi b}+\sin{\pi c}=4\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.

Since $a+b+c=1$, we can set $A=\pi a$, $B=\pi b$, $C=\pi c$ and then have $A+B+C=\pi$. Our goal is to show that

$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$

for any three angles $A$, $B$, $C$ satisfying $A+B+C=\pi$.

Now this can be proven in different ways:

1) One is by writing $C=\pi-A-B$ and simplifying using trigonometric formulae; this is rather boring and it breaks the symmetry.

2) Another one is using complex numbers: let $\alpha=e^{iA/2}$, $\beta=e^{iB/2}$ and $\gamma=e^{iC/2}$. Then,

$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$

becomes

(1) $\dfrac{\alpha^2-\alpha^{-2}}{2i}+\dfrac{\beta^2-\beta^{-2}}{2i}+\dfrac{\gamma^2-\gamma^{-2}}{2i} = 4\cdot\frac{\alpha+\alpha^{-1}}{2}\cdot\frac{\beta+\beta^{-1}}{2}\cdot\frac{\gamma+\gamma^{-1}}{2}$.

Oh, and $A+B+C=\pi$ becomes $\alpha\beta\gamma=2i$. Now proving (1) is just a matter of multiplying out the right hand side and looking at the $8$ terms (two of them, namely $\alpha\beta\gamma$ and $\alpha^{-1}\beta^{-1}\gamma^{-1}$, cancel out, being $i$ and $-i$, respectively).

3) Here is how I would have done it 8 years ago: We can WLOG assume that $A$, $B$, $C$ are the angles of a triangle (this means that $A$, $B$, $C$ lie in the interval $\left[0,\pi\right]$, additionally to satisfying $A+B+C=\pi$), because everything is analytic (or by casebash). We denote the sides of this triangle by $a$, $b$, $c$ (so we forget about the old $a$, $b$, $c$), its semiperimeter $\dfrac{a+b+c}{2}$ by $s$, its area by $\Delta$ and its circumradius by $R$. Then, $\sin A=\dfrac{a}{2R}$ (by the Extended Law of Sines) and similarly $\sin B=\dfrac{b}{2R}$ and $\sin C=\dfrac{c}{2R}$, so that $\sin A+\sin B+\sin C=\dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}=\frac{a+b+c}{2R}=\frac{s}{R}$. On the other hand, one of the half-angle formulas shows that $\cos\dfrac{A}2=\sqrt{\dfrac{s\left(s-a\right)}{bc}}$, and similar formulas hold for $\cos\dfrac{B}2$ and $\cos\dfrac{C}2$, so that

$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$

$=4\cdot\sqrt{\dfrac{s\left(s-a\right)}{bc}}\cdot\sqrt{\dfrac{s\left(s-b\right)}{ca}}\cdot\sqrt{\dfrac{s\left(s-c\right)}{ab}}$

$=\frac{4s}{abc}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.

Now, $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\Delta$ (by Heron's formula) and $\Delta=\frac{abc}{4R}$ (by another formula for the area of the triangle), so tis becomes

$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}=\dfrac{4s}{abc}\cdot\dfrac{abc}{4R}=\dfrac{s}{R}$.

This is exactly what we got for the left hand side, qed.