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

Question

If $a.b,c \in \mathbb{R^+}$ and $ab+bc+ca=1$

Then Prove $$S=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$$

My try we have $$S=\sum \frac{a}{\sqrt{a^2+ab+bc+ca}}=\sum \frac{a}{\sqrt{a+b}\sqrt{a+c})}$$

any hint here?

Answer 1

Not in the spirit you want to g, but here is alternative solution:

Let $x = \arctan {1\over a}$, $y = \arctan {1\over b}$ and $z = \arctan {1\over c}$

Then $a= \cot x$ and so $\frac{a}{\sqrt{a^2+1}} = \cos x\sin x$ ....

Now we have to prove $$\sin 2x+\sin 2y+\sin 2z\leq 3$$ which is true.

Answer 2

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

Answer 3

Let $a=\tan\left(\dfrac{\alpha}{2}\right), b=\tan\left(\dfrac{\beta}{2}\right), a=\tan\left(\dfrac{\gamma}{2}\right) \ \ (\alpha+\beta+\gamma=\pi)$

$$\Leftrightarrow \sin\left(\dfrac{\alpha}{2}\right)+\sin\left(\dfrac{\beta}{2}\right)+\sin\left(\dfrac{\gamma}{2}\right) \le 3\sin\left(\dfrac{\pi}{6}\right)\le \dfrac{3}{2}$$