Inequality with cyclic sum

Question

If $a+b+c=abc$, prove that $\displaystyle \sum_{cyc} \dfrac{1}{\sqrt{a^2+1}} \leqslant \dfrac{3}{2}$.

Answer 1

it is equivalent to $$\cos(x)+\cos(y)+\cos(z)\le \frac{3}{2}$$ which is true for $$x,y,z$$ angles in a triangle

Answer 2

By AM-GM $$\sum_{cyc}\frac{1}{\sqrt{a^2+1}}=\sum_{cyc}\frac{1}{\sqrt{a^2+\frac{abc}{a+b+c}}}=$$ $$=\sum_{cyc}\sqrt{\frac{a+b+c}{a(a^2+ab+ac+bc)}}=\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq$$ $$\leq\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{b+a}\right)=\frac{3}{2}$$