Subfield: Algebra


Question: 设 $x, y, z$ 为正实数, 求 $\left(x+\frac{1}{y}+\sqrt{2}\right)\left(y+\frac{1}{z}+\sqrt{2}\right)\left(z+\frac{1}{x}+\sqrt{2}\right)$ 的最小值.

Solution:
记 $T=\left(x+\frac{1}{y}+\sqrt{2}\right)\left(y+\frac{1}{z}+\sqrt{2}\right)\left(z+\frac{1}{x}+\sqrt{2}\right)$,

当 $x=y=z=1$ 时, $T$ 有最小值 $(2+\sqrt{2})^{3}=20+14 \sqrt{2}$.

下证: $\quad T \geq 20+14 \sqrt{2}$.

 $T=\left(x y z+\frac{1}{x y z}\right)+\sqrt{2}\left(x y+y z+z x+\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\right)$

$$
\begin{aligned}
& +3\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\sqrt{2}\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)+5 \sqrt{2} \\
& \geq 2+\sqrt{2} \times 6 \sqrt[6]{x y \cdot y z \cdot z x \cdot \frac{1}{x y} \cdot \frac{1}{y z} \cdot \frac{1}{z x}} \\
& +3 \times 6 \sqrt[6]{x \cdot y \cdot z \cdot \frac{1}{x} \cdot \frac{1}{y} \cdot \frac{1}{z}}+\sqrt{2} \times 3 \sqrt[3]{\frac{x}{z} \cdot \frac{y}{x} \cdot \frac{z}{y}}+5 \sqrt{2} \\
& =2+6 \sqrt{2}+3 \times 6+\sqrt{2} \times 3+5 \sqrt{2}=20+14 \sqrt{2} .
\end{aligned}
$$

当 $x=y=z=1$ 时, 可取到等号.

所以, $T$ 的最小值为 $20+14 \sqrt{2}$.
记 $T=\left(x+\frac{1}{y}+\sqrt{2}\right)\left(y+\frac{1}{z}+\sqrt{2}\right)\left(z+\frac{1}{x}+\sqrt{2}\right)$,

当 $x=y=z=1$ 时, $T$ 有最小值 $(2+\sqrt{2})^{3}=20+14 \sqrt{2}$.

下证: $\quad T \geq 20+14 \sqrt{2}$.

$$
\quad \begin{aligned}
& T\geq\left(2 \sqrt{\frac{x}{y}}+\sqrt{2}\right)\left(2 \sqrt{\frac{y}{z}}+\sqrt{2}\right)\left(2 \sqrt{\frac{z}{x}}+\sqrt{2}\right) \\
& =8+4 \sqrt{2}\left(\sqrt{\frac{x}{z}}+\sqrt{\frac{z}{y}}+\sqrt{\frac{y}{x}}\right)+4\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)+2 \sqrt{2} \\
& \geq 8+4 \sqrt{2} \times 3 \sqrt[3]{\sqrt{\frac{x}{z}} \cdot \sqrt{\frac{z}{y}} \cdot \sqrt{\frac{y}{x}}}+4 \times 3 \sqrt[3]{\sqrt{\frac{x}{y}} \cdot \sqrt{\frac{y}{z}} \cdot \sqrt{\frac{z}{x}}}+2 \sqrt{2}  \\
& =8+4 \sqrt{2} \times 3+4 \times 3+2 \sqrt{2}=20+14 \sqrt{2} .
\end{aligned}
$$

当 $x=y=z=1$ 时, 可取到等号.

所以, $T$ 的最小值为 $20+14 \sqrt{2}$.
记 $T=\left(x+\frac{1}{y}+\sqrt{2}\right)\left(y+\frac{1}{z}+\sqrt{2}\right)\left(z+\frac{1}{x}+\sqrt{2}\right)$,

当 $x=y=z=1$ 时, $T$ 有最小值 $(2+\sqrt{2})^{3}=20+14 \sqrt{2}$.

下证: $\quad T \geq 20+14 \sqrt{2}$.
注意到: $\frac{x+\frac{1}{y}+\sqrt{2}}{2+\sqrt{2}}=\frac{1}{2+\sqrt{2}} x+\frac{1}{2+\sqrt{2}} \cdot \frac{1}{y}+\frac{\sqrt{2}}{2+\sqrt{2}} \cdot 1 \geq x^{\frac{1}{2+\sqrt{2}}} \cdot\left(\frac{1}{y}\right)^{\frac{1}{2+\sqrt{2}}} \cdot 1^{\frac{\sqrt{2}}{2+\sqrt{2}}}$

于是, $\frac{\left(x+\frac{1}{y}+\sqrt{2}\right)\left(y+\frac{1}{z}+\sqrt{2}\right)\left(z+\frac{1}{x}+\sqrt{2}\right)}{(2+\sqrt{2})^{3}}$

$$
=\frac{x+\frac{1}{y}+\sqrt{2}}{2+\sqrt{2}} \cdot \frac{y+\frac{1}{z}+\sqrt{2}}{2+\sqrt{2}} \cdot \frac{z+\frac{1}{x}+\sqrt{2}}{2+\sqrt{2}}
$$

$$
\begin{aligned}
& \geq\left(x^{\frac{1}{2+\sqrt{2}}} \cdot\left(\frac{1}{y}\right)^{\frac{1}{2+\sqrt{2}}} \cdot 1^{\frac{\sqrt{2}}{2+\sqrt{2}}}\right) \cdot\left(y^{\frac{1}{2+\sqrt{2}}} \cdot\left(\frac{1}{z}\right)^{\frac{1}{2+\sqrt{2}}} \cdot 1^{\frac{\sqrt{2}}{2+\sqrt{2}}}\right) \cdot\left(z^{\frac{1}{2+\sqrt{2}}} \cdot\left(\frac{1}{x}\right)^{\frac{1}{2+\sqrt{2}}} \cdot 1^{\frac{\sqrt{2}}{2+\sqrt{2}}}\right) \\
& =1 .
\end{aligned}
$$

故 $\left(x+\frac{1}{y}+\sqrt{2}\right)\left(y+\frac{1}{z}+\sqrt{2}\right)\left(z+\frac{1}{x}+\sqrt{2}\right) \geq(2+\sqrt{2})^{3}$.

当 $x=y=z=1$ 时, 可取到等号.

所以, $T$ 的最小值为 $(2+\sqrt{2})^{3}=20+14 \sqrt{2}$.

Final answer: $20+14 \sqrt{2}$