设 $x,y,z\in R^+,\sqrt{a}=x(y-z)^2,\sqrt{b}=y(z-x)^2,\sqrt{c}=z(x-y)^2$．求证：$a^2+b^2+c^2\geqslant 2(ab+bc+ca)$．
【难度】
【出处】
2009中国东南数学奥林匹克试题
【标注】
【答案】
略
【解析】
先证 $\sqrt{a},\sqrt{b},\sqrt{c}$ 不能构成三角形的三边．因为
$\begin{aligned}
\sqrt{b}+\sqrt{c}-\sqrt{a}=-(y+z)(z-x)(x-y)\\
\sqrt{c}+\sqrt{a}-\sqrt{b}=-(z+x)(x-y)(y-z)\\
\sqrt{a}+\sqrt{b}-\sqrt{c}=-(x+y)(y-z)(z-x)
\end{aligned}$
所以
$\begin{aligned}&(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}-\sqrt{c})\\
&=-(y+z)(z+x)(x+y)[(y-z)(z-x)(x-y)]^2\\
&=\leqslant 0
\end{aligned}$
于是
$\begin{aligned}
&2(ab+bc+ba)-(a^2+b^2+c^2)\\
&=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}-\sqrt{c})\\
&\leqslant 0
\end{aligned}$
故 $a^2+b^2+c^2\geqslant 2(ab+bc+ca)$．
$\begin{aligned}
\sqrt{b}+\sqrt{c}-\sqrt{a}=-(y+z)(z-x)(x-y)\\
\sqrt{c}+\sqrt{a}-\sqrt{b}=-(z+x)(x-y)(y-z)\\
\sqrt{a}+\sqrt{b}-\sqrt{c}=-(x+y)(y-z)(z-x)
\end{aligned}$
所以
$\begin{aligned}&(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}-\sqrt{c})\\
&=-(y+z)(z+x)(x+y)[(y-z)(z-x)(x-y)]^2\\
&=\leqslant 0
\end{aligned}$
于是
$\begin{aligned}
&2(ab+bc+ba)-(a^2+b^2+c^2)\\
&=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}-\sqrt{c})\\
&\leqslant 0
\end{aligned}$
故 $a^2+b^2+c^2\geqslant 2(ab+bc+ca)$．
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