2. Let $a, b, c \geqslant 1$, and positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
a^{x}+b^{y}+c^{z}=4, \\
x a^{x}+y b^{y}+z x^{z}=6, \\
x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z}=9 .
\end{array}\right.
$$

Then the maximum possible value of $c$ is $\qquad$
2. $\sqrt[3]{4}$.

By Cauchy-Schwarz inequality
$$
\begin{array}{l}
\left(a^{x}+b^{y}+c^{z}\right)\left(x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z}\right) \\
\geqslant\left(x a^{x}+y b^{y}+z b^{z}\right)^{2},
\end{array}
$$

we know that equality holds only when $x=y=z$.
Then $x\left(a^{x}+b^{x}+c^{x}\right)=6 \Rightarrow x=\frac{3}{2}$.
Thus, when $a=b=1$, $c$ attains its maximum value $\sqrt[3]{4}$.