一道有趣的不等式题

已知 \(x 、 y 、 z\) 是正数，且 \(x y z=1,\) 求\(P\)的最小值:
\(P=2(x+y+z)+\frac{x}{y^{3}+z^{3}+1}+\frac{y}{z^{3}+x^{3}+1}+\frac{z}{x^{3}+y^{3}+1}\)

先把后三项拿出来

\(\begin{aligned} & \frac{x}{y^{3} + z^{3}+1}+\frac{y}{z^{3}+x^{3}+1}+\frac{z}{x^{3}+y^{3}+1}\\ =& \frac{x^{2}}{x\left(y^{3}+z^{3}\right)+x}+\frac{y^{2}}{y\left(z^{3}+x^{3}\right)+y}+\frac{z^{2}}{z\left(x^{3}+y^{3}\right)+z} \\\geqslant &\frac{(x+y+z)^{2}}{x\left(y^{3}+z^{3}\right)+y\left(z^{3}+x^{3}\right)+z\left(x^{3}+y^{3}\right)+(x+y+z)} \\= & \frac{(x+y+z)^{2}}{x y\left(x^{2}+y^{2}\right)+y z\left(y^{2}+z^2\right)+ z x\left(z^{2}+x^{2}\right)+\left(x+y+z\right)x y z}\\=& \frac{(x+y+z)^{2}}{\left(x^{2}+y^{2}+z^{2}\right)(x y+y z+z x)} \\=&\frac{(x+y+z)^{2}(x y+y z+z x)}{\left(x^{2}+y^{2}+z^{2}\right)(x y+y z+z x)(x y+y z+z x)} \end{aligned}\)

其中除了用到了\(x y z=1\)进行代换，还用到了不等式\(\sum\frac{a_i^2}{a_ib_i}\geqslant\frac{\left(\sum a_i\right)^2}{\sum a_ib_ i}\)(将分母乘到左边，柯西不等式可证)

然后使用均值不等式

\(\begin{aligned} & \frac{(x+y+z)^{2}(x y+y z+z x)}{\left(x^{2}+y^{2}+z^{2}\right)(x y+y z+z x)(x y+y z+z x)} \\\geqslant& \frac{(x+y+z)^{2} \cdot 3 \sqrt[3]{x y \cdot y z\cdot zx}}{\left(\frac{(x+y+z)^{2}}{3}\right)^{3}} \\=&\frac{81}{(x+y+z)^{4}} \end{aligned}\)

\(\begin{aligned} \therefore P &=2(x+y+z)+\frac{x}{y^{3}+z^{3}+1}+\frac{y}{z^{3}+x^{3}+1}+\frac{z}{x^{3}+y^{3}+1} \\ & \geqslant 2(x+y+z)+\frac{81}{(x+y+z)^{4}} \\ &=4 \left (\frac{x+y+z}{3}\right)+\frac{81}{(x+y+z)^{4}}+\frac{2}{3}(x+y+z) \\ &\geqslant 5 \sqrt[5]{\left(\frac{x+y+z}{3}\right)^{4} \cdot \frac{81}{(x+y+z)^{4}}}+\frac{2}{3} \cdot 3 \sqrt[3]{x y z}\\&=5+2=7 \end{aligned}\)

最后这一步比较妙，注意到第一次使用均值不等式时取等条件为\(x=y=z=1\)，后续使用不等式时必须满足此取等条件，因此这里将\(2(x+y+z)\)拆分成\(5\)项而非简单的\(4\)项