Maximum value of a expression Problem: If $\alpha+\beta+\gamma=20$,  then what is $\max(\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5})$?
My attempt: Assume $\alpha \geq \beta \geq \gamma$. Then $\alpha+\beta+\gamma \leq 3\alpha$ so $25 \leq 3\alpha+5$.
Also $\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\leq 3\sqrt{3\alpha+5}$
At this point, I don't have idea what to do next. What should I do next?
Am I doing it incorrectly?

Use Cauchy-Bunyakovsky-Schwarz inequality:
$$\left(\sum_{i=1}^3u_iv_i\right)^2\le \left(\sum_{i=1}^3u_i^2\right)\left(\sum_{i=1}^3v_i^2\right)$$
We choose $u_1=\sqrt{3\alpha+5}$ and so on, and $v_i=1$. Then $$\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\le\sqrt{(3\alpha+5+3\beta+5+3\gamma+5)3}=15$$
The equality is achieved when all the terms are equal, for $\alpha=\beta=\gamma=20/3$.
