<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] Change the notation using  $x=\frac{1}{a}$ , $y=\frac{1}{b}$ and $z=\frac{1}{c}$
Now the condition is that $xy+yz+zx=1$.
And the inequality to prove is :
$$\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}} \leq \frac{3}{2}$$
Now use the condition and notice that :
$$x^2+1=x^2+xy+yz+zx=(x+y)(x+z)$$
So now using the AM-GM inequality :
$$\frac{x}{\sqrt{x^2+1}}=\frac{x}{\sqrt{(x+y)(x+z)}} =\sqrt{\frac{x^2}{(x+y)(x+z)}} \leq \frac{1}{2} \left (\frac{x}{x+y}+\frac{x}{x+z} \right )$$
Do the same thing for the other terms and add them to get the conclusion .
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