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?
 A: 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 $\alpha=\beta=\gamma=20/3$.
A: Let $u=\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}$.
Denote:
$$\begin{cases}
3\alpha +5=a^2\\
3\beta+5=b^2\\
3\gamma +5=c^2
\end{cases}$$
Then, the problem becomes:
$$v=a+b+c\to max \ \ s.t.\ \ a^2+b^2+c^2=75$$and:
$$v^2=a^2+b^2+c^2+2(ab+bc+ca)=75+2(ab+bc+ca)\le \\
75+2(a^2+b^2+c^2)=225,$$
the equality occurs for $a=b=c=5$.
Hence: $v(5,5,5)=15$ or $u(20/3,20/3,20/3)=15$ is max.
A: $\alpha+\beta+\gamma=20$
$\implies (3 \alpha + 5) + (3 \beta + 5) + (3 \gamma + 5) = 75$
By AM-QM inequality,
$\cfrac {\sqrt{3 \alpha + 5} + \sqrt{3 \beta + 5} + \sqrt{3 \gamma + 5}}{3} \leq \sqrt{\cfrac{(3 \alpha + 5) + (3 \beta + 5) + (3 \gamma + 5)}{3}}$
leads to $\sqrt{3 \alpha + 5} + \sqrt{3 \beta + 5} + \sqrt{3 \gamma + 5} \leq 15$
Equality occurs at $\alpha = \beta = \gamma = \cfrac{20}{3}$
