问题 \(a,b,c \ge 0\), 求证:
\[
\begin{aligned}
&\sqrt{a^2 - ab + b^2} \cdot \sqrt{b^2 - bc + c^2} \\
&{}+ \sqrt{b^2 - bc + c^2} \cdot \sqrt{c^2 - ca + a^2} \\
&{}+ \sqrt{c^2 - ca + a^2} \cdot \sqrt{a^2 - ab + b^2}
\ge a^2 + b^2 + c^2
\end{aligned}
\]

Qer: mawentao 20170914

解答 (By Tr. J. R. Guo)

\[
\begin{aligned}
&
\sum \sqrt{a^2 - ab + b^2} \cdot \sqrt{b^2 - bc + c^2} \\
{}={}&
\sum \sqrt{ \left( a - \dfrac{b}2\right)^2 + \dfrac34 b^2 } \cdot \sqrt{\left( c - \dfrac{b}2\right)^2 + \dfrac34 b^2 } \\
{}\ge{}&
\sum \sqrt{\left[ \left( a - \dfrac{b}2\right) \cdot \left( c - \dfrac{b}2\right) + \dfrac34 b^2 \right]^2} \\
{}={} &
\sum \left| ac - \dfrac12 ab - \dfrac12 bc + b^2 \right| \\
{} \ge {}&
\sum \left( ac - \dfrac12 ab - \dfrac12 bc + b^2 \right) = \sum b^2
\end{aligned}
\]