<s>[INST] Solve $\sqrt{\frac{(2 - x)(2 - y)}{(2 + x)(2 + y)} + \frac{(2 - y)(2 - z)}{(2 + y)(2 + z)} + \frac{(2 - z)(2 - x)}{(2 + z)(2 + x)}}$ 
Find the value of the expression $S$, knowing that $x^2 + y^2 + z^2 + xyz = 4$ and $x, y, z \in (0, \infty)$:

$$
S = \sqrt{\frac{(2 - x)(2 - y)}{(2 + x)(2 + y)} + \frac{(2 - y)(2 - z)}{(2 + y)(2 + z)} + \frac{(2 - z)(2 - x)}{(2 + z)(2 + x)}}
$$
My attempt:
As $x^2 + y^2 + z^2 + xyz = 4$, we can choose $A, B, C \in \left[0, \frac{\pi}{2}\right)$ for which $x = 2 \cos A$, $y = 2 \cos B$ și $z = 2 \cos C$, a well-known substitution identity.
Now, let's prove that:
$$\tan^2 A = \frac{2 - x}{2 + x}$$
It is true that:
$$\cos A = \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}}$$
Therefore:
$$\tan^2 A = \frac{1 - \cos A}{1 + \cos A}$$
Equivalent to the identity we wanted to prove true.
Let's substitute in the expression:
$$S = \sqrt{\tan^2 \frac{A}{2} \tan^2 \frac{B}{2} + \tan^2 \frac{B}{2} \tan^2 \frac{C}{2} + \tan^2 \frac{C}{2} \tan^2 \frac{A}{2}}$$
Which doesn't seem to be constant. I suspect the problem is actually wrong, because if, instead of one square root, we'd have three square roots, the experession would have evaluated to:
$$S'= \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1$$
Which is calculable by a very well known identity. Could the problem be wrong? Or it's something I'm missing, a further step to solve the initial problem? Thanks in advance!
 [/INST] One cannot find the value of $S$, only simplified expression.
Let's denote $x+y+z=a$, $xy+xz+yz=b$, $xyz=c$.
$x^2+y^2+z^2=a^2-2b \Rightarrow a^2-2b+c=4 \Rightarrow 2b=a^2+c-4$.
$S^2=A/B$, $A=(2-x)(2-y)(2+z)+(2-x)(2-z)(2+y)+(2-y)(2-z)(2+x)=$
$3xyz-2(xy+yz+xz)-4(x+y+z)+24=3c-2b-4a+24$,
$B=(2+x)(2+y)(2+z)=xyz+2(xy+yz+xz)+4(x+y+z)+8=c+2b+4a+8$.
Using $2b=a^2+c-4$ one can obtain $A=3c-a^2-c+4-4a+24=2c+32-(a+2)^2$,
$B=c+a^2+c-4+4a+8=2c+(a+2)^2$.
So, the answer is $S=\sqrt{\frac{2c+32-d}{2c+d}}$, where $c=xyz$,
$d=(x+y+z+2)^2$.
The value of $S$ is not constant. $x=2,y=z\approx 0 \Rightarrow S\approx 1$; $x=y=z=1 
\Rightarrow S=\sqrt{\frac{1}{3}}$.
Maybe, the word "minimum" or "maximum" is lost before "value".
$S^2=\frac{2c+32-d}{2c+d}$. At constant $c$ $S^2$ is decreasing with
growth of $d$, so the maximum $S$ is corresponding with minimum $d$ and
vice versa.
Minimum $S^2 \to 23-16\sqrt{2}$ is the limit at $x \to 0$ and $y=z 
=\sqrt{\frac{4}{x+2}}$. Maximum $S^2\to 1$ is the limit at $x \to 0$ and
$y=x$, $z$ is positive root of $z^2+x^2 z+2 x^2=4$.
Maybe, the original problem should be "to prove that $S<1$". Using
$S^2=\frac{2c+32-d}{2c+d}$, it is equivalent to $d=(x+y+z+2)^2>16 
\Leftrightarrow x+y+z>2$.
Using that "geometric mean is not greater than quadratic mean" one can
write $\sqrt[3]{c}=\sqrt[3]{xyz}\leq\sqrt{\frac{x^2+y^2+z^2}{3}}= 
\sqrt{\frac{4-xyz}{3}}=\sqrt{\frac{4-c}{3}} \Rightarrow 27c^2\leq 
(4-c)^3 \Rightarrow c \leq 1$.
Using that "garmonic mean is not greater than geometic mean" one can
write $\frac{3c}{b}=\frac{1}{\frac{1}{3}\cdot 
\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)} \leq \sqrt[3]{xyz}= 
\sqrt[3]{c} \Rightarrow$ $\frac{27c^3}{b^3}\leq c \Rightarrow$
$b^3\geq{27c^2} \Rightarrow b\geq 3 \sqrt[3]{c^2} \geq 3c$ for $c\leq 1$. Using it in $a^2-2b+c=4 \Rightarrow$ $a^2=4+2b-c \geq 4+6c-c > 4$ for $c > 0$. So, $a^2 > 4 \Rightarrow a>2 
\Rightarrow d>16 \Rightarrow S^2 < 1 \Rightarrow S < 1$.
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