An inequality, which is supposed to be simple Let $x,y,z\in\mathbb{R}$.Let $xy+yz+xz=1$.
Prove:$\displaystyle \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\leq \frac{3}{2}$

Hint:  $$x^2 + 1 = x^2 + xy + yz + zx = (x+y)(x+z)$$
Also, you will need the inequality
$$
\sqrt{AB}\le \frac12(A+B)
$$

Solution:
$$ \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}
\\=  \frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(x+z)}}
+\frac{z}{\sqrt{(x+y)(x+z)}}
\\
\le \frac12\left[
\frac x{x+y} + \frac x{x+z} + 
\frac y{y+z} + \frac y{x+y} + 
\frac z{z+y} + \frac z{z+x} 
\right]=3/2
$$
