A gamma function identity In some of my previous work on mean values of Dirichlet L-functions, I came upon the following identity for the Gamma function:
\begin{equation} 
\frac{\Gamma(a) \Gamma(1-a-b)}{\Gamma(1-b)} 
+ \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} 
+ \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} = 
\pi^{\frac12} 
\frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)}
\frac{\Gamma\left(\frac{a}{2}\right)}{\Gamma\left(\frac{1-a}{2}\right)}
\frac{\Gamma\left(\frac{b}{2}\right)}{\Gamma\left(\frac{1-b}{2}\right)}.
\end{equation}
As is often the case, once one knows such a formula should be true then it is easy to prove it.  I give my proof below.  My questions are 1) Has this formula been observed before?  I have no idea how to search the literature for such a thing.  2) Is there a better proof? (Of course this is totally subjective, but one thing that would please me would be to avoid trigonometric functions since they do not appear in the formula.)
Proof.
Using
\begin{equation}
\frac{\Gamma(\frac{s}{2})}{\Gamma(\frac{1-s}{2})} = \pi^{-\frac12} 2^{1-s} \cos({\textstyle \frac{\pi s}{2}}) \Gamma(s),
\end{equation}
the right hand side is
\begin{equation}
2 \frac{\cos(\frac{\pi a}{2}) \cos(\frac{\pi b}{2}) \Gamma(a) \Gamma(b)}{\cos(\frac{\pi (a + b)}{2}) \Gamma(a+b)}.
\end{equation}
On the other hand, the left hand side is
\begin{equation}
\frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} 
\left(
 \frac{\Gamma(a+b) \Gamma(1-a-b)}{\Gamma(b) \Gamma(1-b)} 
 + \frac{\Gamma(a+b) \Gamma(1-a-b)}{\Gamma(a) \Gamma(1-a)}
 + 1
\right),
\end{equation}
which becomes after using $\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)}$,
\begin{equation}
\frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} 
\left(\frac{\sin(\pi a) + \sin( \pi b) + \sin(\pi(a + b))}{\sin(\pi(a+b))} \right).
\end{equation}
Using trig formulas, we get that this is
\begin{equation}
 2 \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} \frac{\sin(\frac{\pi}{2}(a+b)) \cos(\frac{\pi}{2}(a-b)) + \sin(\frac{\pi}{2}(a+b)) \cos(\frac{\pi}{2}(a+b)) }{\sin(\pi(a+b))}
\end{equation}
I think I've run out of space?  The rest is easy trig.
 A: I decided to take the opposite approach to what Matt Y. asked for (sorry!), which was to see exactly how his proof depends on trig function identities.  As it turns out, you can use trig identities to produce additional Gamma function identities like the original formula, though usually without being quite as clean and symmetric.
To see how it goes, let's look at Matt's proof.  The classical reflection and doubling formulas for the the Gamma function produce the formulas used in the proof:
$$
\Gamma(s)\sin(\pi s) = \frac{\pi}{\Gamma(1-s)}
$$
and 
$$
\Gamma(s)\cos \Bigl( \frac{\pi s}{2}\Bigr) = \pi^{1/2} 2^{s-1} \frac{\Gamma \bigl(
\frac{s}{2} \bigr)}{\Gamma \bigl( \frac{1-s}{2} \bigr)}.
$$
Consider the trig identity,
$$
  \frac{ \sin(x) + \sin(y) }{\sin(x+y)}  + 1 = \frac{2\cos \bigl(\frac{x}{2} \bigr) \cos \bigl( \frac{y}{2} \bigr)}{\cos \bigl( \frac{x+y}{2} \bigr)},
$$
which is easy enough to verify (and is floating around in the proof already).  If we take $x=\pi a$ and $y=\pi b$ and multiply this identity by $\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$, then the previous $\Gamma$-function identities yield Matt's formula.
So I wondered how easy it would be to use known trig identities to produce additional formulas of this type, say with different denominators, and here is an example.  Again using the reflection and multiplication formulas, you can show that
$$
\Gamma(s)\biggl( \frac{1}{2} + \cos\biggl(\frac{ 2\pi s}{3} \biggr) \biggr) = \pi \frac{3^{s-1/2} \Gamma\bigl(\frac{s}{3} \bigr)}{\Gamma \bigl( \frac{1-s}{3} \bigr) \Gamma \bigl( \frac{2-s}{3} \bigr)}.
$$
The trig identity I want to use (coming from the triple angle formulas for cosine) is
$$
  2 \biggl( \frac{1}{2} + \cos\biggl( \frac{2x}{3} \biggr) \biggr)^3 - 3 \biggl( \frac{1}{2} + \cos \biggl( \frac{2x}{3} \biggr) \biggr)^2 + \sin^2(x) = 0.
$$
Take $x=\pi s$ and multiply by $\Gamma(s)^3$.  Then after the dust settles this yields the identity
$$
2\pi 3^{3s-3/2}\frac{ \Gamma\bigl( \frac{s}{3} \bigr)^3}{ \Gamma\bigl( \frac{1-s}{3} \bigr)^3 \Gamma \bigl( \frac{2-s}{3} \bigr)^3} - 3^{2s} \frac{ \Gamma(s) \Gamma\bigl( \frac{s}{3} \bigr)^2}{\Gamma \bigl( \frac{1-s}{3} \bigr)^2 \Gamma \bigl( \frac{2-s}{3} \bigr)^2} + \frac{ \Gamma(s)}{\Gamma(1-s)^2} = 0.
$$
So yeah, it's not that pretty and we've only obtained a one-variable identity, but presumably one can find more examples of multivariable formulas in a similar fashion --- though you need to get the stars to align properly.  What is nice about Matt Y.'s formula is that the powers of 2 that could be there (much like the powers of 3 here) nicely cancel away, leaving only $\Gamma$-values and a single power of $\pi$.
A: This is not yet an answer, but in terms of $B$ (beta) functions, one has
$B(a,1-a-b)+B(b,1-a-b)+B(a,b)=\displaystyle\frac{B\left(\frac{b}{2},\frac{1}{2}\right)}{B\left(\frac{1-a}{2},\frac{a+b}{2}\right)}$
(use the fact that $\Gamma(1/2)=\sqrt{\pi})$. 
Now, i think one can use the additive properties that beta functions enjoy such as $B(a,b)=B(a+1,b)+B(a,b+1)$. hmm....
A: (I'm adding this as an answer since my original post won't let me add any more info, and the formula is too big for a comment.)
There is a variant on the formula as follows
\begin{equation}
\frac{\Gamma(a ) \Gamma(1-a-b)}{ \Gamma(1-b)} 
+ \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} 
- \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} =
\pi^{\frac12} 
\frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)}
\frac{\Gamma\left(\frac{1+a}{2}\right)}{\Gamma\left(\frac{2-a}{2}\right)}
\frac{\Gamma\left(\frac{1+b}{2}\right)}{\Gamma\left(\frac{2-b}{2}\right)}.
\end{equation}
By adding/subtracting it with the original formula one gets an identity with the sum of two terms on  each side.  I didn't find it helpful but perhaps it sparks an idea with someone.
In the context of my original problem, the identity with + corresponds to a sum over even Dirichlet characters, while the identity in this answer corresponds to odd characters.  The gamma factors on the right hand side are relevant for the functional equation of the Dirichlet L-functions which depends on the parity of the character.
A: 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) $\dfrac{\alpha^2-\alpha^{-2}}{2i}+\dfrac{\beta^2-\beta^{-2}}{2i}+\dfrac{\gamma^2-\gamma^{-2}}{2i} = 4\cdot\dfrac{\alpha+\alpha^{-1}}{2}\cdot\dfrac{\beta+\beta^{-1}}{2}\cdot\dfrac{\gamma+\gamma^{-1}}{2}$.
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 $\sin A+\sin B+\sin C=\dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}=\dfrac{a+b+c}{2R}=\dfrac{s}{R}$. 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
$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$
$=4\cdot\sqrt{\dfrac{s\left(s-a\right)}{bc}}\cdot\sqrt{\dfrac{s\left(s-b\right)}{ca}}\cdot\sqrt{\dfrac{s\left(s-c\right)}{ab}}$
$=\dfrac{4s}{abc}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.
Now, $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\Delta$ (by Heron's formula) and $\Delta=\dfrac{abc}{4R}$ (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.
A: A proof of the statement has already been given, so I will just add a small historical remark.
The left hand side of your identity is the Veneziano amplitude in the case of four identical scalar particles. The right hand side corresponds to another crossing symmetric four-point amplitude for scalar particles suggested by Virasoro in "Alternative Constructions of Crossing-Symmetric Amplitudes with Regge Behavior". These are known to coincide for the choice of parameters $a+b+c=1$ giving your identity. I would recommend the article "The birth of string theory" for a quick survey of the Veneziano model and its generalization to N scalar particles and related mathematics.
A: Jon Borwein recently told a class that the way to prove identities involving $\Gamma$ is to write the purported identity in the form $\Gamma(a)={\rm whatever}$ and then show that the right side satisfies the hypotheses of Bohr-Mollerup, that is, it's 1 when $a=1$, it satisfies $f(x+1)=xf(x)$, and it's log-convex. I have no idea whether that will work with this particular question, but maybe it's worth a try. 
EDIT: It occurs to me that everything in the identity can be written in terms of Beta functions. Maybe it's easier to find if you look for it as a Beta-function identity. 
