Inequality (4)

$\underline{\textit{First Solution:}}$

Set $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z} \implies xy+yz+zx=1$ and the inequality: $\sum \frac{x}{\sqrt{x^2+1}}\le \frac{3}{2}$ ,

But $\sum \frac{x}{\sqrt{x^2+1}} = \sum \frac{x}{\sqrt{x^2+xy+xz+zy}} = \sum \sqrt{\frac{x}{x+y}\frac{x}{x+z}}$ ,

Also, $\sum \sqrt{\frac{x}{x+y}\frac{x}{x+z}} \le \frac{\frac{x}{x+y}+\frac{x}{x+z}}{2}$ ,

Thus, $\sum \sqrt{\frac{x}{x+y}\frac{x}{x+z}} \le \frac{\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{x+y} +\frac{y}{y+z}+\frac{z}{z+y}+\frac{z}{z+x}}{2} = \frac{3}{2}$ ,

$\underline{\textit{Second Solution:}}$

Subsitute $a=\tan x, b=\tan y, c=\tan z \text{ where } x+y+z=\pi$ ,

then $\cos x+\cos y+\cos z \le \frac{3}{2}$

$\underline{\textit{Third Solution:}}$

By AM-GM we have $a+b+c \ge 3\sqrt[3]{abc} \text { and since } a+b+c = abc \implies (abc)^2 \ge 27$ ,

We can rewrite the inequality as: $\frac{1}{3}$ $(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}) \le \frac{1}{2}$ ,

since $f(a)=\frac{1}{\sqrt{a^2+1}}$ is concave, we apply Jensen's inequality:

$\frac{1}{3} f(a)+ \frac{1}{3} f(b) + \frac{1}{3} f(c) \le f( \frac{a+b+c}{3}) = f(\frac{abc}{3})= \frac{1}{\sqrt{\frac{(abc)^2}{3^2}+1}} \le \frac{1}{2} \iff (abc)^2\ge 27$