Given two positive reals $x$, $y$, I want to find the strictest upper bound to $\sqrt{x}\sqrt{y}$ in the form $h(kx + g(y))$. where $k$ must be a real and $g$ and $h$ can be any functions from $\mathbb{R}$ to $\mathbb{R}$.
It is pretty straightforward to show $$ \sqrt{x}\sqrt{y} \le \min_{\alpha>0}\left( \frac{\alpha}{\sqrt{2}}x + \frac{1}{\alpha\sqrt{2}}y\right)\enspace. $$
Is it possible to find stricter bounds?
Any reference would also be appreciated.
we have $$x\geq 0$$ and $$y\geq 0$$ thus we get after AM-GM: $$\sqrt{x}\sqrt{y}\le \frac{1}{2}(x+y)$$