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$\tan(\alpha+\beta+\gamma)=\tan\frac{\pi}2$ $\implies \dfrac{\displaystyle\sum_{\text{cyc}}\tan \alpha-\displaystyle\prod_{\text{cyc}}\tan \alpha}{1-\displaystyle\sum_{\text{cyc}}\tan \alpha\tan \beta}=\tan\frac{\pi}2\to \infty$

For this to hold true: $\color{cyan}{\displaystyle\sum_{\text{cyc}}\tan \alpha\tan \beta=1}$ .

Now from adding the given conditions:

$\color{#D61F06}{A+B+C}= \color{cyan}{\displaystyle\sum_{\text{cyc}}\tan \alpha\tan \beta}\color{#333333}{+15=}\color{#D61F06}{16}$

Using Cauchy Shwarz inequality ,

$\sqrt A+\sqrt B+\sqrt C\le \sqrt{(\color{#D61F06}{A+B+C})\color{#333333}{(1+1+1)}}=\sqrt{48}\approx{6.982}$

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Equality occurs when : $\small{A=B=C\text{ OR }\alpha=\beta=\gamma=\dfrac{\pi}6}$