Problem
Consider three positive numbers $a$, $b$ and $c$. As the function $x \mapsto \sqrt{x}$ is concave, for any $(\alpha, \beta, \gamma)$ with $\alpha + \beta + \gamma = 1$ and $\alpha,\, \beta,\, \gamma \ge 0$: \begin{equation} \alpha \sqrt{a} + \beta \sqrt{b} + \gamma \sqrt{c} \le \sqrt{\alpha a + \beta b + \gamma c}. \end{equation}
Can we find other families of upperbound? By "family of upperbound" I mean a set of points $u = f(a,b,c, \alpha, \beta, \gamma)$ linear in $a$, $b$ and $c$ such that $\alpha \sqrt{a} + \beta \sqrt{b} + \gamma \sqrt{c} \le \sqrt{u}$.
Intuition
The convex combination $\alpha \sqrt{a} + \beta \sqrt{b} + \gamma \sqrt{c} $ may be view as a convex combination of two convex comibations: \begin{equation} \theta \left(\omega_{A} \sqrt{a} + (1-\omega_{A}) \sqrt{c}\right) + (1-\theta) \left(\omega_{B} \sqrt{b} + (1-\omega_{B}) \sqrt{c}\right), \end{equation} with $\omega_A, \, \omega_B, \, \theta \in [0,1]$.
Can't we use some property of the derivative $x\mapsto \sqrt{x}$ to produce better upperbound than just taking twice the concavity property?
I would like to add second order terms in the expression of $u$, e.g., $(\alpha^2 - \beta\gamma)a$.