<s>[INST] Three Variables-Inequality with $a+b+c=abc$ $a$,$b$,$c$ are positive numbers such that $~a+b+c=abc$
Find the maximum value of $~\dfrac{1}{\sqrt{1+a^{2}}}+\dfrac{1}{\sqrt{1+b^{2}}}+\dfrac{1}{\sqrt{1+c^{2}}}$
 [/INST] Trigonometric substitution looks good for this, especially if you know sum of cosines of angles in a triangle are $\le \frac32$.  However if you want an alternate way...
Let $a = \frac1x, b = \frac1y, c = \frac1z$.  Then we need to find the maximum of
$$F = \sum \frac{x}{\sqrt{x^2+1}}$$
with the condition now as $xy + yz + zx = 1$.  Using this condition to homogenise, we have
$$\begin{align}
F &= \sum \frac{x}{\sqrt{x^2 + xy + yz + zx}} \\
&= \sum \frac{x}{\sqrt{(x+y)(x+z)}} \\
&= \sum \frac{x\sqrt{(x+y)(x+z)}}{(x+y)(y+z)} \\
&\le \sum \frac{x\left((x+y)+(x+z)\right)}{2\cdot(x+y)(y+z)} \quad \text{by AM-GM}\\
&= \frac12 \sum \left(\frac{x}{x+y}+\frac{x}{x+z} \right) \\
&= \frac12 \sum \left(\frac{x}{x+y}+\frac{y}{y+x} \right) \quad \text{rearranging terms across sums}\\
&= \frac32
\end{align}$$
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