How could you solve this statement :
$1 - \frac{1}{2} I_{\frac{a}{a+x^2}}(\frac{a}{2},\frac{1}{2}) = \frac{\sqrt{a}}{2} (\sqrt{I_{\frac{x}{x+1}}(\frac{a}{2},\frac{a}{2})} - \frac{1}{\sqrt{I_{\frac{x}{x+1}}(\frac{a}{2},\frac{a}{2})}}) $
with Hint :
$I_x(a,a) = 1 - \frac{1}{2}I_{4x(1-x)}(a,\frac{1}{2}) ,$ where 1/2<= x <= 1 .
Some definitions : $I_x(a,b) = \frac{B(x;a,b)}{B(a,b)} $ ,where B(x;a,b) is incomplete beta function and B(a,b) is beta function.
I tried to set left side minus right side to be a function f. Then I considered $\frac{df}{dx}$ and f(1), if both are equal $0$ then it is proved. But it doesn't go any further. Could anyone help me ?
As writen, I have the strange feeling that this is not an identity.
Consider the function $$f(x,a)=\text{lhs - rhs}$$ Consider the constant term in the series expansion of $f(x,a)$ around $x=1$. For $a=1$, it is $$\frac{\left(4+\sqrt{2}\right) \pi -2 B_{\frac{1}{2}}\left(\frac{1}{2},\frac{1}{2}\right)}{4 \pi }\neq 0$$ and, beside this specific value, it is an increasing function of $a$.
Doing the same around $x=\frac 12$, the constant term of the expansion is an increasing function of $a$ and, for $a=1$, this constant is $\neq 0$.
Now, making explicit the function, we have $$f(x,2)=\frac{1}{2} \left(\sqrt{\frac{x^2}{x^2+2}}+\frac{\sqrt{2}}{\sqrt{x} \sqrt{x+1}}+1\right)$$
Similarly, $f'(1,a)$ is not a constant. It is $0$ only for $a=1$.