The Cauchy-Schwarz inequality states that given 2 sequences of $n$ real numbers $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$, then $\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)\left(y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}\right) \geq\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{n} y_{n}\right)^{2}$ with equality holding if and only if $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=\ldots=\frac{x_{n}}{y_{n}}$. Applying this to $\left\{a^{x / 2}, b^{y / 2}, c^{z / 2}\right\}$ and $\left\{x a^{x / 2}, y b^{y / 2}, z c^{z / 2}\right\}$ yields $\left(a^{x}+b^{y}+c^{z}\right)\left(x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z}\right) \geq\left(x a^{x}+y b^{y}+z b^{z}\right)^{2}$ with equality holding if and only if $x=y=z$. However, equality does hold (both sides evaluate to 36), so $x=y=z$. The second equation then becomes $x\left(a^{x}+b^{x}+c^{x}\right)=6$, which implies $x=\frac{3}{2}$. Then we have $a^{3 / 2}+b^{3 / 2}+c^{3 / 2}=4$. To maximize $c$, we minimize $a$ and $b$ by setting $a=b=1$. Then $c^{3 / 2}=2$ or $c=\sqrt[3]{4}$.