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
(egin{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}}{xleft(y^{3}+z^{3} ight)+x}+frac{y^{2}}{yleft(z^{3}+x^{3} ight)+y}+frac{z^{2}}{zleft(x^{3}+y^{3} ight)+z} \geqslant &frac{(x+y+z)^{2}}{xleft(y^{3}+z^{3} ight)+yleft(z^{3}+x^{3} ight)+zleft(x^{3}+y^{3} ight)+(x+y+z)} \= & frac{(x+y+z)^{2}}{x yleft(x^{2}+y^{2} ight)+y zleft(y^{2}+z^2 ight)+ z xleft(z^{2}+x^{2} ight)+left(x+y+z ight)x y z}\=& frac{(x+y+z)^{2}}{left(x^{2}+y^{2}+z^{2} ight)(x y+y z+z x)} \=&frac{(x+y+z)^{2}(x y+y z+z x)}{left(x^{2}+y^{2}+z^{2} ight)(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(sumfrac{a_i^2}{a_ib_i}geqslantfrac{left(sum a_i ight)^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
(egin{aligned} & frac{(x+y+z)^{2}(x y+y z+z x)}{left(x^{2}+y^{2}+z^{2} ight)(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 zcdot zx}}{left(frac{(x+y+z)^{2}}{3} ight)^{3}} \=&frac{81}{(x+y+z)^{4}} end{aligned})
(egin{aligned} herefore 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} ight)+frac{81}{(x+y+z)^{4}}+frac{2}{3}(x+y+z) \ &geqslant 5 sqrt[5]{left(frac{x+y+z}{3} ight)^{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