Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$

Question

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.

Answer 1

By AM-GM $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sum\limits_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3}{2}$.

Answer 2

Set $a=\cot A$ etc.

$\implies\sum\cot A\cot B=1$

$\iff\tan A+\tan B+\tan C=\tan A\tan B\tan C$

$\implies A+B+C=n\pi$ where $n$ is any integer

Now use In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$

Answer 3

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.