设 $a、b、c$ 是正实数，求证：$\dfrac{a}{\sqrt{{{a}^{2}}+8bc}}+\dfrac{b}{\sqrt{{{b}^{2}}+8ca}}+\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}\geqslant 1$．（韩国）
【难度】
【出处】
2001年第42届IMO试题
【标注】
【答案】
略
【解析】
证法一
我们先证明
$\dfrac{a}{\sqrt{a^2+8bc}}\geqslant \dfrac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$
即 $(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})^2\geqslant a^{\frac{2}{3}}(a^2+8bc)$．
由于平均不等式
$\begin{aligned}
&(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})^2-(a^{\frac{4}{3}})^2\\
&=(b^{\frac{4}{3}}+c^{\frac{4}{3}})(a^{\frac{4}{3}}+a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})\\
&\geqslant 2b^{\frac{2}{3}}c^{\frac{2}{3}}\cdot 4a^{\frac{2}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}\\
&=8a^{\frac{2}{3}}bc
\end{aligned}$
所以
$(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})^2\geqslant (a^{\frac{4}{3}})^2+8a^{\frac{2}{3}}bc=a^{\frac{2}{3}}(a^2+8bc)$
所以 $\dfrac{a}{\sqrt{a^2}+8bc}\geqslant \dfrac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$．
同样地，有 $\dfrac{b}{\sqrt{{{b}^{2}}+8ca}}\geqslant\dfrac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}},\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}\geqslant \dfrac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$
把三个不等式相加，便得 $\dfrac{a}{\sqrt{{{a}^{2}}+8bc}}+\dfrac{b}{\sqrt{{{b}^{2}}+8ca}}+\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}\geqslant 1$．
证法二
令 $x=\dfrac{a}{\sqrt{{{a}^{2}}+8bc}},y=\dfrac{b}{\sqrt{{{b}^{2}}+8ca}},z=\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}$，则 $x,y,z$ 是正实数，且 $x^2=\dfrac{a^2}{a^2+8bc}$，于是 $\dfrac{1}{x^2}-1=\dfrac{8bc}{a^2}$．
同样地，$\dfrac{1}{y^2}-1=\dfrac{8ca}{b^2},\dfrac{1}{z^2}=\dfrac{8ab}{c^2}$．
所以 $\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{y^2}-1\right)\left(\dfrac{1}{z^2}-1\right)=512$．
若 $x+y+z<1$，则 $0<x<1,0<y<1,0<z<1$，于是
$\begin{aligned}
&\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{x^2}-1\right)\\
&=\frac{(1-x^2)(1-y^2)(1-z^2)}{x^2y^2z^2}\\
&>\dfrac{[(x+y+z)^2-x^2][(x+y+z)^2-y^2][(x+y+z)^2-z^2]}{x^2y^2z^2}\\
&=\dfrac{(y+z)(2x+y+z)(z+x)(2y+z+x)(x+y)(2z+x+y)}{x^2y^2z^2}\\
&\geqslant\dfrac{2\sqrt{yz}\cdot 4\sqrt[4]{x^2yz}\cdot 2\sqrt{zx}\cdot 4\sqrt[4]{y^2zx}\cdot 2\sqrt{xy}\cdot 4\sqrt[4]{z^2xy}}{x^2y^2z^2}\\
&=\dfrac{512x^2y^2z^2}{x^2y^2z^2}=512
\end{aligned}$
矛盾．
所以 $x+y+z\geqslant 1$，即 $\dfrac{a}{\sqrt{{{a}^{2}}+8bc}}+\dfrac{b}{\sqrt{{{b}^{2}}+8ca}}+\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}\geqslant 1$．
我们先证明
$\dfrac{a}{\sqrt{a^2+8bc}}\geqslant \dfrac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$
即 $(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})^2\geqslant a^{\frac{2}{3}}(a^2+8bc)$．
由于平均不等式
$\begin{aligned}
&(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})^2-(a^{\frac{4}{3}})^2\\
&=(b^{\frac{4}{3}}+c^{\frac{4}{3}})(a^{\frac{4}{3}}+a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})\\
&\geqslant 2b^{\frac{2}{3}}c^{\frac{2}{3}}\cdot 4a^{\frac{2}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}\\
&=8a^{\frac{2}{3}}bc
\end{aligned}$
所以
$(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}})^2\geqslant (a^{\frac{4}{3}})^2+8a^{\frac{2}{3}}bc=a^{\frac{2}{3}}(a^2+8bc)$
所以 $\dfrac{a}{\sqrt{a^2}+8bc}\geqslant \dfrac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$．
同样地，有 $\dfrac{b}{\sqrt{{{b}^{2}}+8ca}}\geqslant\dfrac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}},\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}\geqslant \dfrac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$
把三个不等式相加，便得 $\dfrac{a}{\sqrt{{{a}^{2}}+8bc}}+\dfrac{b}{\sqrt{{{b}^{2}}+8ca}}+\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}\geqslant 1$．
证法二
令 $x=\dfrac{a}{\sqrt{{{a}^{2}}+8bc}},y=\dfrac{b}{\sqrt{{{b}^{2}}+8ca}},z=\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}$，则 $x,y,z$ 是正实数，且 $x^2=\dfrac{a^2}{a^2+8bc}$，于是 $\dfrac{1}{x^2}-1=\dfrac{8bc}{a^2}$．
同样地，$\dfrac{1}{y^2}-1=\dfrac{8ca}{b^2},\dfrac{1}{z^2}=\dfrac{8ab}{c^2}$．
所以 $\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{y^2}-1\right)\left(\dfrac{1}{z^2}-1\right)=512$．
若 $x+y+z<1$，则 $0<x<1,0<y<1,0<z<1$，于是
$\begin{aligned}
&\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{x^2}-1\right)\\
&=\frac{(1-x^2)(1-y^2)(1-z^2)}{x^2y^2z^2}\\
&>\dfrac{[(x+y+z)^2-x^2][(x+y+z)^2-y^2][(x+y+z)^2-z^2]}{x^2y^2z^2}\\
&=\dfrac{(y+z)(2x+y+z)(z+x)(2y+z+x)(x+y)(2z+x+y)}{x^2y^2z^2}\\
&\geqslant\dfrac{2\sqrt{yz}\cdot 4\sqrt[4]{x^2yz}\cdot 2\sqrt{zx}\cdot 4\sqrt[4]{y^2zx}\cdot 2\sqrt{xy}\cdot 4\sqrt[4]{z^2xy}}{x^2y^2z^2}\\
&=\dfrac{512x^2y^2z^2}{x^2y^2z^2}=512
\end{aligned}$
矛盾．
所以 $x+y+z\geqslant 1$，即 $\dfrac{a}{\sqrt{{{a}^{2}}+8bc}}+\dfrac{b}{\sqrt{{{b}^{2}}+8ca}}+\dfrac{c}{\sqrt{{{c}^{2}}+8ab}}\geqslant 1$．
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