If \( a+b+c=0 \), then prove that \[\frac{(b+c)^{2}}{3 b c}+\frac{(c+a)^{2}}{3 a c}+\frac{(a+b)^...

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If \( a+b+c=0 \), then prove that
\[\frac{(b+c)^{2}}{3 b c}+\frac{(c+a)^{2}}{3 a c}+\frac{(a+b)^{2}}{3 a b}=1\]
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