<s>[INST] Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}.$ 
Let $a,b,$ and $c$ be positive real numbers such that $a+b+c = abc$. Prove that $$\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}.$$

This question seems tricky because how do we incorporate the $a+b+c = abc$ condition to solve this? I was thinking of solving for the variables and substituting like $a = \dfrac{b+c}{1-bc}$ etc., but I am not sure how that will help.
 [/INST] Let $a=\tan(A)$, $b=\tan(B)$ and $c=\tan(C)$. We then have $$\tan(A)+\tan(B)+\tan(C) = \tan(A)\tan(B)\tan(C) \,\,\,\, (\spadesuit)$$
Recall that $(\spadesuit)$ is true iff $A+B+C = \pi$, i.e., $A$, $B$ and $C$ are angles of a triangle. (Infact the part we need is trivial, since we have $\tan(C) = -\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} = \tan(\pi-(A+B))$.)
Further, since $a,b,c > 0$, we have that $0 \leq A,B,C < \dfrac{\pi}2$. We then have that
$$\dfrac1{\sqrt{1+a^2}} + \dfrac1{\sqrt{1+b^2}} + \dfrac1{{\sqrt{1+c^2}}} = \underbrace{\cos(A) + \cos(B) + \cos(C) \leq 3\cos\left
(\dfrac{A+B+C}3\right)}_{\text{By Jensen}} = \dfrac32$$
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