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

Attempt
The $ab+bc+ca = 1$ condition reminds of the rearrangement inequality. Thus, I would say that $a^2+b^2+c^2 \geq ab+bc+ca = 1$ then rewrite the given inequality as $4(a+b+c)^2 = 4(a^2+b^2+c^2) + 8(ab+bc+ca) = 4(a^2+b^2+c^2) \leq 9(a^2+1)(b^2+1)(c^2+1)$
I don't know what to do next.
 [/INST] See my previous answer here: How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$
For an elegant solution, apply the so-called Purkiss Principle.
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