Is there any way to show that the following inequality holds for the given functions with constraints?
Let $f_1(x) = \frac{1}{2}(\sqrt{(1+(1+a) x+(1-a) y)^2-4 a x (1+x)}+(1-a)x-(1-a)y-1)$,
$f_2(x) = \frac{1}{2}(\sqrt{(1+(1+a) x+(1-a) y)^2-4 a x (1+x)}-(1-a)x+(1-a)y-1)$, and
$f_3(x) = \frac{(1-a)(y+1)x}{(1-a)y+1}$, such that
$\frac{\left(1+\frac{1}{f_1(x)}\right)^{f'_1(x)}\left(1+\frac{1}{f_2(x)}\right)^{f'_2(x)}}{\left(1+\frac{1}{f_3(x)}\right)^{f'_3(x)}} \geq 1$, for $0.5 \leq a \leq 1$, $x >0,y \geq 0$. It can be easily checked that it saturates the inequality for $\lim_{x \rightarrow \infty}$.
Here is the full expression:
$\frac{\left(1+\frac{2}{\left(\sqrt{(1+(1+a) x+(1-a) y)^2-4 a x (1+x)}+(1-a)x-(1-a)y-1\right)}\right)^{\frac{(1 - a) (1 + (1+a) y + (1-a)x)}{2\sqrt{(1+(1+a) x+(1-a) y)^2-4 a x (1+x)}}+\frac{(1-a)}{2}}\left(1+\frac{2}{\left(\sqrt{(1+(1+a) x+(1-a) y)^2-4 a x (1+x)}-(1-a)x+(1-a)y-1\right)}\right)^{\frac{(1 - a) (1 +(1+a) y + (1-a)x)}{2\sqrt{(1+(1+a) x+(1-a) y)^2-4 a x (1+x)}}-\frac{(1-a)}{2}}}{\left(1+\frac{(1-a)y+1}{(1-a)(y+1)x}\right)^{\frac{(1-a)(y+1)}{(1-a)y+1}}} \geq 1$..
I have noticed the following points: Each $f_i(x) > 0 $ for the given constraints! Hence all individual terms inside the big brackets are greater than 1. Moreover, the term written in the power of each bracket is always $\geq 0$.
Numerically, I've checked this for different range of $x,y$ and $a$. It always holds.. I'm not sure how I can prove this analytically! Actual expression was in terms of $\log$ functions and it was required to show that the $\log$ function is always positive $\rightarrow$ the above mentioned function is always $\geq 1$.
I am sorry for such a complex form of the expression. Thanks!