Problem 3.5. Find all triples $(x, y, z)$ of real numbers that are solutions to the system of equations

$$
\left.\begin{array}{l}
3 \cdot 2^{y}-1=2^{x}+2^{-x} \\
3 \cdot 2^{z}-1=2^{y}+2^{-y} \\
3 \cdot 2^{x}-1=2^{z}+2^{-z}
\end{array}\right\}
$$
Solution Problem 3.5. By making the substitution $2^{x}=a, 2^{y}=b$, and $2^{z}=c$, it is observed that $a, b, c>0$ and we obtain

$$
\left.\begin{array}{l}
b=\frac{1}{3}\left(a+1+\frac{1}{a}\right) \\
c=\frac{1}{3}\left(b+1+\frac{1}{b}\right) \\
a=\frac{1}{3}\left(c+1+\frac{1}{c}\right)
\end{array}\right\}
$$

Applying the inequality between the arithmetic and geometric means, we get

$$
b=\frac{1}{3}\left(a+1+\frac{1}{a}\right) \geq \sqrt[3]{a \cdot 1 \cdot \frac{1}{a}}=1
$$

and thus $b \geq 1 / b$. Similarly, $a \geq 1, a \geq 1 / a$, and $c \geq 1, c \geq 1 / c$. Taking the above into account, from the first equation we have

$$
b=\frac{1}{3}\left(a+1+\frac{1}{a}\right) \leq \frac{1}{3}(a+a+a)=a
$$

and from the other two that $c \leq b$ and $a \leq c$. Combining the above inequalities, we get $a \leq c \leq b \leq a$. That is, $a=b=c$.

Now we have

$$
a=\frac{1}{3}\left(a+1+\frac{1}{a}\right) \Leftrightarrow 2 a^{2}-a-1=(a-1)(2 a+1)=0
$$

which has the solution $a=1$ and $a=-1 / 2$. Since only the positive solution is valid, we have $a=b=c=1$ and therefore, $(x, y, z)=(0,0,0)$ is the only triplet solution of the system.

Another Solution Problem 3.5. By making the substitution $2^{x}=a, 2^{y}=b$, and $2^{z}=c$, it is observed that $a, b, c>0, \mathrm{and}$ we obtain:

$$
\left.\left.\left.\begin{array}{l}
3 b-1=a+\frac{1}{a}, \\
3 c-1=b+\frac{1}{b}, \\
3 a-1=c+\frac{1}{c} .
\end{array}\right\} \Leftrightarrow \quad \begin{array}{l}
a(3 b-1)=a^{2}+1 \\
b(3 c-1)=b^{2}+1 \\
c(3 a-1)=c^{2}+1
\end{array}\right\} \Leftrightarrow \quad \begin{array}{l}
3 a b=a^{2}+a+1 \\
3 b c=b^{2}+b+1 \\
3 c a=c^{2}+c+1
\end{array}\right\}
$$

Observe that since $a^{2}-2 a+1=(a-1)^{2} \geq 0$, we have $a^{2}+1 \geq 2 a$, so the first equation implies $3 a b \geq 3 a$, hence $b \geq 1$. Similarly, from the other two equations we have $c \geq 1$ and $a \geq 1$.

Suppose that $a$ is the smallest of the three values $a, b, c$ (the other cases are analogous). In particular, $0<a \leq b$. Then we have $a^{2}+a+1=3 a b \geq 3 a^{2}$, so $2 a^{2} \leq a+1$. The function $2 x^{2}$ can only be less than or equal to the function $x+1$ if $x \leq 1$. Therefore, $a \leq 1$, and thus $a=1$.

The first equation then becomes $3 b=3$, so $b=1$. And from the second equation, we get $c=1$. Therefore, $2^{x}=2^{y}=2^{z}=1$, and thus the only possible triplet is $(x, y, z)=(0,0,0)$.