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Posted: **Sat Jan 24, 2009 5:57 pm**

Aratati ca daca \( a,\ b,\ c\in \mathb{R} \) si \( a+b+c=6 \), atunci \( a^2+b^2+c^2\ge 12 \).

Petrisor Neagoe, Etapa locala, Bihor 2006

Posted: **Sat Jan 24, 2009 9:02 pm**

\( \sum a^2\geq\frac{(a+b+c)^2}{3}=12. \)

OLM Bihor 2006