An inequality, which is supposed to be simple Let $x,y,z\in\mathbb{R}$.Let $xy+yz+xz=1$.
Prove:$\displaystyle \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\leq \frac{3}{2}$
 A: 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:
$$ \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}
\\=  \frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(x+z)}}
+\frac{z}{\sqrt{(x+y)(x+z)}}
\\
\le \frac12\left[
\frac x{x+y} + \frac x{x+z} + 
\frac y{y+z} + \frac y{x+y} + 
\frac z{z+y} + \frac z{z+x} 
\right]=3/2
$$
A: Let $a,b,c\in(0,\pi)$ be such that $x=\cot a$, $y=\cot b$, $z=\cot c$.
Using the addition formula for cotangent, one can show that
$$ \cot(a+b+c)
= \frac{\cot a\cot b\cot c - \cot a - \cot b - \cot c}
    {\cot a\cot b + \cot b\cot c + \cot c\cot a - 1}
   \tag{$\ast$} $$
By hypothesis, the denominator on the RHS is 0, so (see below) the LHS is $\infty$; since $a,b,c\in (0,\pi)$, this implies $a+b+c\in\{\pi,2\pi\}$.
Case $a+b+c=\pi$: Then at most one of $a,b,c$ is greater than $\frac\pi2$; wlog, $a,b\in(0,\frac\pi2]$.  Since cosine is concave on that interval, we have
\begin{align*}
\frac{x}{\sqrt{x^2+1}} + \frac{y}{\sqrt{y^2+1}} + \frac{z}{\sqrt{z^2+1}}
&= \cos a + \cos b + \cos c \\
&\le 2\cos(\tfrac{a+b}{2}) + \cos c \\
&= 2\cos(\tfrac{a+b}{2}) - \cos(a+b) \\
&= 2\cos(\tfrac{a+b}{2}) - 2\cos^2(\tfrac{a+b}{2}) + 1 \\
&= \tfrac32 - 2\big(\cos(\tfrac{a+b}{2}) - \tfrac12\big)^2 \\
&\le \tfrac32
\end{align*}
with equality iff $a=b=\frac\pi3$ (whence $c=\frac\pi3$ also), that is, $x=y=z=\frac1{\sqrt3}$.
Case $a+b+c=2\pi$: Then at most one of $a,b,c$ is less than $\frac\pi2$; wlog, $a,b\in[\frac\pi2,\pi)$.  Since cosine is nonpositive in that interval, we have
$$ \frac{x}{\sqrt{x^2+1}} + \frac{y}{\sqrt{y^2+1}} + \frac{z}{\sqrt{z^2+1}}
= \cos a + \cos b + \cos c
\le \cos c
\le 1
< \tfrac32
$$

Now, one annoying detail about ($\ast$): I noted that the denominator on the RHS is zero, and inferred that the RHS, and hence the LHS, is $\infty$.  But what if the numerator on the RHS is also zero?  Well, suppose for contradiction that both numerator and denominator are zero.  Then $xyz=x+y+z$ and $xy+yz+zx=1$, and so
\begin{align*}
(t-x)(t-y)(t-z)
&= t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz \\
&= t^3 - xyzt^2 + t - xyz \\
&= (t^2+1)(t-xyz)
\end{align*}
which is impossible because the first polynomial has three real roots (counting multiplicity) but the last has only one.
