I'm trying to prove that $u_1(x) \geq u_2(x)$ for $ x \in [0,1] $.
$$u_1(x) = \frac{1}{2} \cdot h^{-1}\left(\frac{1}{2} \cdot h(v(1)) + \frac{1}{2} \cdot h(v(x))\right) + \frac{1}{2} \cdot h^{-1}\left(\frac{1}{2} \cdot h(v(0)) + \frac{1}{2} \cdot h(v(1 - x))\right)$$
$$u_2(x) = h^{-1}\left(\frac{1}{2} \cdot h\left(\frac{1}{2} \cdot v(1) + \frac{1}{2} \cdot v(1 - x)\right) + \frac{1}{2} \cdot h\left(\frac{1}{2} \cdot v(x) + \frac{1}{2} \cdot v(0)\right)\right)$$
Both $v(x)$ and $h(x)$ are concave, and increasing. I've plugged in many different functional forms and always found that $u_1(x) \geq u_2(x)$ but I'm struggling to come up with an analytical proof. Even with the assumption that $v(x)$ and $h(x)$ are continuously differentiable (which is an assumption I'm cool with making), I'm not able to make any progress.
Any help would be greatly appreciated! Thank you!
P.S. It's readily shown that $u_1(0)=u_2(0)$ and $u_1(1)=u_2(1)$ and both are maximised at least at some $x>0.5$.