<s>[INST] 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?
 [/INST] $\alpha+\beta+\gamma=20$
$\implies (3 \alpha + 5) + (3 \beta + 5) + (3 \gamma + 5) = 75$
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 $\alpha = \beta = \gamma = \cfrac{20}{3}$
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