I have the following function and want to find the upper bound of this function.
$$\max_{p\in (0, \,1/2]} \left(\frac{\sqrt{2}-1}{\sqrt{2}-2^{\frac{2p-1}{2p}}}\right)^p$$ The plot of this function shows that it is quadratic (convex) and the maximum of this function is at $p=0$ and $p=0.5$, which is same and equal to 1. But how can I reach to this upper bound mathematically. If you could suggest me some steps.
According to the suggestion of Vincent 7, I did the following work
$$\log(f)=\max_{p\in (0, \,1/2]} p\log\left(\frac{\sqrt{2}-1}{\sqrt{2}-2^{\frac{2p-1}{2p}}}\right)$$
$$\log(f)=\max_{p\in (0, \,1/2]} p\,\left(\log(\sqrt{2}-1)-\log\left(\sqrt{2}-2^{\frac{2p-1}{2p}}\right)\right)$$
$$\log(f)=\max_{p\in (0, \,1/2]} p\,\left(-0.88-\log\left(\sqrt{2}-2^{\frac{2p-1}{2p}}\right)\right)$$
$$\log(f)=\max_{p\in (0, \,1/2]} - p\,\left(0.88+\log\left(\sqrt{2}-2^{\frac{2p-1}{2p}}\right)\right)$$
$$f=\max_{p\in (0, \,1/2]} e^{- p\,\left(0.88+\log\left(\sqrt{2}-2^{\frac{2p-1}{2p}}\right)\right)}$$
and this also leads to the quadratic convex form. How we can get upper bound, I am not getting yet. Any suggestion???