正实数 $x、y、z$ 满足 $xyz\geqslant 1$，证明：$\dfrac{{{x}^{5}}-{{x}^{2}}}{{{x}^{5}}+{{y}^{2}}+{{z}^{2}}}+\dfrac{{{y}^{5}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{5}}+{{z}^{2}}}+\dfrac{{{z}^{5}}-{{z}^{2}}}{{{x}^{2}}+{{y}^{2}}+{{z}^{5}}}\geqslant 0$．（韩国）
【难度】
【出处】
2005年第46届IMO试题
【标注】
【答案】
略
【解析】
证法一
我们只需证明：$\displaystyle \sum\dfrac{x^5}{x^5+y^2+z^2}\geqslant 1\geqslant \sum\dfrac{x^2}{x^5+y^2+z^2}$ ①
设 $xyz=d^3\geqslant 1$，令 $x=x_1d,y=y_1d,z=z_1d$，则 $x_1y_1z_1=1$，且
$\displaystyle \begin{aligned}
\sum\frac{x^5}{x^5+y^2+z^2}&=\sum\frac{x_1^5d^3}{x_1^5d^3+y_1^2+z_1^2}\\\
&=\sum\frac{x_1^5}{x_1^5+\frac{1}{d^3}(y_1^2+z_1^2)}\\
&\geqslant \sum\frac{x_1^5}{x_1^5+y_1^2+z_1^2}
\end{aligned}$
$\displaystyle \begin{aligned}
\sum\frac{x^2}{x^5+y^2+z^2}&=\sum\frac{x_1^2d^2}{x_1^5d^5+y_1^2d^2+z_1^2d^2}\\\
&=\sum\frac{x_1^2}{x_1^5d^3+y_1^2+z_1^2}\\
&\leqslant \sum\frac{x_1^2}{x_1^5+y_1^2+z_1^2}
\end{aligned}$
所以，我们只需在 $xyz=1$ 的情况下，证明 ① 式．
因为
$\displaystyle \begin{aligned}
\sum\frac{x^5}{x^5+y^2+z^2}&=\sum\frac{x^5}{x^5+xyz(y^2+z^2)}\\
&=\sum\frac{x^4}{x^4+y^3z+yz^3}\\
&\geqslant\sum\frac{x^4}{x^4+y^4+z^4}=1
\end{aligned}$
所以，① 的左边得证．
而 $\displaystyle \sum\dfrac{x^2}{x^5+y^2+z^2}=\sum\dfrac{x^2\cdot xyz}{x^5+xyz(y^2+z^2)}=\sum\dfrac{x^2yz}{x^4+yz(y^2+z^2)}$
由平均不等式
$\begin{aligned}
x^4+x^4+y^3z+yz^3&\geqslant 4x^2yz\\
x^4+y^3z+y^3z+y^2z^2&\geqslant 4xy^2z\\
x^4+yz^3+yz^3+y^2z^2&\geqslant 4xyz^2\\
y^3z+yz^3&\geqslant 2y^2z^2
\end{aligned}$
把上面 $4$ 个不等式相加，可得 $x^4+yz(y^2+z^2)\geqslant x^2yz+xy^2z+xyz^2$
所以 $\displaystyle \sum\dfrac{x^2}{x^5+y^2+z^2}=\sum\dfrac{x^2yz}{x^4+y^3z+yz^3}\leqslant\sum\dfrac{x^2yz}{x^2yz+xy^2z+xyz^2}=1$
从而 ① 的右边得证．
证法二
因为 $\dfrac{x^5-x^2}{x^5+y^2+z^2}-\dfrac{x^5-x^2}{x^3(x^2+y^2+z^2)}=\dfrac{x^2(x^3-1)^2(y^2+z^2)}{x^3(x^5+y^2+z^2)(x^2+y^2+z^2)}\geqslant 0$
所以
$\displaystyle \begin{aligned}
\sum\frac{x^5-x^2}{x^5+y^2+z^2}&\geqslant\frac{x^5-x^2}{x^3(x^2+y^2+z^2)}\\
&=\frac{1}{x^2+y^2+z^2}\sum\left(x^2-\frac{1}{x}\right)\\
&\geqslant\frac{1}{x^2+y^2+z^2}\sum(x^2-yz)(因为xyz\geqslant 1)\\
&\geqslant 0
\end{aligned}$
我们只需证明：$\displaystyle \sum\dfrac{x^5}{x^5+y^2+z^2}\geqslant 1\geqslant \sum\dfrac{x^2}{x^5+y^2+z^2}$ ①
设 $xyz=d^3\geqslant 1$，令 $x=x_1d,y=y_1d,z=z_1d$，则 $x_1y_1z_1=1$，且
$\displaystyle \begin{aligned}
\sum\frac{x^5}{x^5+y^2+z^2}&=\sum\frac{x_1^5d^3}{x_1^5d^3+y_1^2+z_1^2}\\\
&=\sum\frac{x_1^5}{x_1^5+\frac{1}{d^3}(y_1^2+z_1^2)}\\
&\geqslant \sum\frac{x_1^5}{x_1^5+y_1^2+z_1^2}
\end{aligned}$
$\displaystyle \begin{aligned}
\sum\frac{x^2}{x^5+y^2+z^2}&=\sum\frac{x_1^2d^2}{x_1^5d^5+y_1^2d^2+z_1^2d^2}\\\
&=\sum\frac{x_1^2}{x_1^5d^3+y_1^2+z_1^2}\\
&\leqslant \sum\frac{x_1^2}{x_1^5+y_1^2+z_1^2}
\end{aligned}$
所以，我们只需在 $xyz=1$ 的情况下，证明 ① 式．
因为
$\displaystyle \begin{aligned}
\sum\frac{x^5}{x^5+y^2+z^2}&=\sum\frac{x^5}{x^5+xyz(y^2+z^2)}\\
&=\sum\frac{x^4}{x^4+y^3z+yz^3}\\
&\geqslant\sum\frac{x^4}{x^4+y^4+z^4}=1
\end{aligned}$
所以，① 的左边得证．
而 $\displaystyle \sum\dfrac{x^2}{x^5+y^2+z^2}=\sum\dfrac{x^2\cdot xyz}{x^5+xyz(y^2+z^2)}=\sum\dfrac{x^2yz}{x^4+yz(y^2+z^2)}$
由平均不等式
$\begin{aligned}
x^4+x^4+y^3z+yz^3&\geqslant 4x^2yz\\
x^4+y^3z+y^3z+y^2z^2&\geqslant 4xy^2z\\
x^4+yz^3+yz^3+y^2z^2&\geqslant 4xyz^2\\
y^3z+yz^3&\geqslant 2y^2z^2
\end{aligned}$
把上面 $4$ 个不等式相加，可得 $x^4+yz(y^2+z^2)\geqslant x^2yz+xy^2z+xyz^2$
所以 $\displaystyle \sum\dfrac{x^2}{x^5+y^2+z^2}=\sum\dfrac{x^2yz}{x^4+y^3z+yz^3}\leqslant\sum\dfrac{x^2yz}{x^2yz+xy^2z+xyz^2}=1$
从而 ① 的右边得证．
证法二
因为 $\dfrac{x^5-x^2}{x^5+y^2+z^2}-\dfrac{x^5-x^2}{x^3(x^2+y^2+z^2)}=\dfrac{x^2(x^3-1)^2(y^2+z^2)}{x^3(x^5+y^2+z^2)(x^2+y^2+z^2)}\geqslant 0$
所以
$\displaystyle \begin{aligned}
\sum\frac{x^5-x^2}{x^5+y^2+z^2}&\geqslant\frac{x^5-x^2}{x^3(x^2+y^2+z^2)}\\
&=\frac{1}{x^2+y^2+z^2}\sum\left(x^2-\frac{1}{x}\right)\\
&\geqslant\frac{1}{x^2+y^2+z^2}\sum(x^2-yz)(因为xyz\geqslant 1)\\
&\geqslant 0
\end{aligned}$
答案
解析
备注
