设 $x,y,z$ 为正实数，求 $(x+\dfrac{1}{y}+\sqrt{2})(y+\dfrac{1}{z}+\sqrt{2})(z+\dfrac{1}{x}+\sqrt{2})$ 的最小值．
【难度】
【出处】
2018年全国高中数学联赛四川省预赛
【标注】
【答案】
$20+14\sqrt{2}$
【解析】
记 $T=(x+\dfrac{1}{y}+\sqrt{2})(y+\dfrac{1}{z}+\sqrt{2})(z+\dfrac{1}{x}+\sqrt{2})$，当 $x=y=z=1$ 时，$T$ 有最小值 $(2+\sqrt{2})^3=20+14\sqrt{2}$．下证：$T\geqslant 20+14\sqrt{2}$．
解法一 $T=(xyz+\dfrac{1}{xyz})+\sqrt{2}(xy+yz+zx+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx})+3(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})+\sqrt{2}(\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y})+5\sqrt{2}\\
\geqslant 2+\sqrt{2}\times 6\sqrt[6]{xy\cdot yz\cdot zx\cdot \dfrac{1}{xy}\cdot\dfrac{1}{yz}\cdot\dfrac{1}{zx}}+3\times 6\sqrt[6]{x\cdot y\cdot z\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\dfrac{1}{z}}+\sqrt{2}\times 3\sqrt[3]{\dfrac{x}{z}\cdot\dfrac{y}{x}\cdot\dfrac{z}{y}}+5\sqrt{2}\\
=2+6\sqrt{2}+3\times 6+\sqrt{2}\times 3+5\sqrt{2}=20+14\sqrt{2}$．当 $x=y=z=1$ 时，可取到等号．所以，$T$ 的最小值为 $20+14\sqrt{2}$．
解法二
$T\geqslant \left( 2\sqrt{\dfrac{x}{y}}+\sqrt{2}\right)\left( 2\sqrt{\dfrac{y}{z}}+\sqrt{2} \right)\left( 2\sqrt{\dfrac{z}{x}}+\sqrt{2}\right)\\
=8+4\sqrt{2}\left( \sqrt{\dfrac{x}{z}}+\sqrt{\dfrac{z}{y}}+\sqrt{\dfrac{y}{x}}\right)+4\left( \sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{z}}+\sqrt{\dfrac{z}{x}}\right)+2\sqrt{2}\\
\geqslant 8+4\sqrt{2}\times 3\sqrt[3]{\sqrt{\dfrac{x}{z}}\cdot \sqrt{\dfrac{z}{y}}\cdot\sqrt{\dfrac{y}{x}}}+4\times 3\sqrt[3]{\sqrt{\dfrac{x}{y}}\cdot \sqrt{\dfrac{y}{z}}\cdot\sqrt{\dfrac{z}{x}}}+2\sqrt{2}\\
=8+4\sqrt{2}\times 3+4\times3+2\sqrt{2}=20+14\sqrt{2}$
当 $x=y=z=1$ 时，可取到等号．所以，$T$ 的最小值为 $20+14\sqrt{2}$．
解法三
注意到 $\dfrac{x+\dfrac{1}{y}+\sqrt{2}}{2+\sqrt{2}}=\dfrac{1}{2+\sqrt{2}}x+\dfrac{1}{2+\sqrt{2}}\cdot\dfrac{1}{y}+\dfrac{\sqrt{2}}{2+\sqrt{2}}\cdot 1\geqslant {{x}^{\dfrac{1}{2+\sqrt{2}}}}\cdot{{\left( \dfrac{1}{y} \right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}}$．于是，$\dfrac{\left(x+\dfrac{1}{y}+\sqrt{2} \right)\left( y+\dfrac{1}{z}+\sqrt{2} \right)\left( z+\dfrac{1}{x}+\sqrt{2}\right)}{{{\left( 2+\sqrt{2} \right)}^{3}}}\\
=\dfrac{x+\dfrac{1}{y}+\sqrt{2}}{2+\sqrt{2}}\cdot\dfrac{y+\dfrac{1}{z}+\sqrt{2}}{2+\sqrt{2}}\cdot \dfrac{z+\dfrac{1}{x}+\sqrt{2}}{2+\sqrt{2}}\\
\geqslant\left[ {{x}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{\left( \dfrac{1}{y} \right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot{{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}} \right]\cdot \left[ {{y}^{\dfrac{1}{2+\sqrt{2}}}}\cdot{{\left( \dfrac{1}{z} \right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}}\right]\cdot \left[ {{z}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{\left( \dfrac{1}{x}\right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}}\right]=1$．故 $(x+\dfrac{1}{y}+\sqrt{2})(y+\dfrac{1}{z}+\sqrt{2})(z+\dfrac{1}{x}+\sqrt{2})\geqslant (2+\sqrt{2})^3$．当 $x=y=z=1$ 时，可取到等号．所以，$T$ 的最小值为 $20+14\sqrt{2}$．所以，$T$ 的最小值为 $(2+\sqrt{2})^3=20+14\sqrt{2}$．
解法一 $T=(xyz+\dfrac{1}{xyz})+\sqrt{2}(xy+yz+zx+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx})+3(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})+\sqrt{2}(\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y})+5\sqrt{2}\\
\geqslant 2+\sqrt{2}\times 6\sqrt[6]{xy\cdot yz\cdot zx\cdot \dfrac{1}{xy}\cdot\dfrac{1}{yz}\cdot\dfrac{1}{zx}}+3\times 6\sqrt[6]{x\cdot y\cdot z\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\dfrac{1}{z}}+\sqrt{2}\times 3\sqrt[3]{\dfrac{x}{z}\cdot\dfrac{y}{x}\cdot\dfrac{z}{y}}+5\sqrt{2}\\
=2+6\sqrt{2}+3\times 6+\sqrt{2}\times 3+5\sqrt{2}=20+14\sqrt{2}$．当 $x=y=z=1$ 时，可取到等号．所以，$T$ 的最小值为 $20+14\sqrt{2}$．
解法二
$T\geqslant \left( 2\sqrt{\dfrac{x}{y}}+\sqrt{2}\right)\left( 2\sqrt{\dfrac{y}{z}}+\sqrt{2} \right)\left( 2\sqrt{\dfrac{z}{x}}+\sqrt{2}\right)\\
=8+4\sqrt{2}\left( \sqrt{\dfrac{x}{z}}+\sqrt{\dfrac{z}{y}}+\sqrt{\dfrac{y}{x}}\right)+4\left( \sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{z}}+\sqrt{\dfrac{z}{x}}\right)+2\sqrt{2}\\
\geqslant 8+4\sqrt{2}\times 3\sqrt[3]{\sqrt{\dfrac{x}{z}}\cdot \sqrt{\dfrac{z}{y}}\cdot\sqrt{\dfrac{y}{x}}}+4\times 3\sqrt[3]{\sqrt{\dfrac{x}{y}}\cdot \sqrt{\dfrac{y}{z}}\cdot\sqrt{\dfrac{z}{x}}}+2\sqrt{2}\\
=8+4\sqrt{2}\times 3+4\times3+2\sqrt{2}=20+14\sqrt{2}$
当 $x=y=z=1$ 时，可取到等号．所以，$T$ 的最小值为 $20+14\sqrt{2}$．
解法三
注意到 $\dfrac{x+\dfrac{1}{y}+\sqrt{2}}{2+\sqrt{2}}=\dfrac{1}{2+\sqrt{2}}x+\dfrac{1}{2+\sqrt{2}}\cdot\dfrac{1}{y}+\dfrac{\sqrt{2}}{2+\sqrt{2}}\cdot 1\geqslant {{x}^{\dfrac{1}{2+\sqrt{2}}}}\cdot{{\left( \dfrac{1}{y} \right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}}$．于是，$\dfrac{\left(x+\dfrac{1}{y}+\sqrt{2} \right)\left( y+\dfrac{1}{z}+\sqrt{2} \right)\left( z+\dfrac{1}{x}+\sqrt{2}\right)}{{{\left( 2+\sqrt{2} \right)}^{3}}}\\
=\dfrac{x+\dfrac{1}{y}+\sqrt{2}}{2+\sqrt{2}}\cdot\dfrac{y+\dfrac{1}{z}+\sqrt{2}}{2+\sqrt{2}}\cdot \dfrac{z+\dfrac{1}{x}+\sqrt{2}}{2+\sqrt{2}}\\
\geqslant\left[ {{x}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{\left( \dfrac{1}{y} \right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot{{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}} \right]\cdot \left[ {{y}^{\dfrac{1}{2+\sqrt{2}}}}\cdot{{\left( \dfrac{1}{z} \right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}}\right]\cdot \left[ {{z}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{\left( \dfrac{1}{x}\right)}^{\dfrac{1}{2+\sqrt{2}}}}\cdot {{1}^{\dfrac{\sqrt{2}}{2+\sqrt{2}}}}\right]=1$．故 $(x+\dfrac{1}{y}+\sqrt{2})(y+\dfrac{1}{z}+\sqrt{2})(z+\dfrac{1}{x}+\sqrt{2})\geqslant (2+\sqrt{2})^3$．当 $x=y=z=1$ 时，可取到等号．所以，$T$ 的最小值为 $20+14\sqrt{2}$．所以，$T$ 的最小值为 $(2+\sqrt{2})^3=20+14\sqrt{2}$．
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