Concursul "Nicolae Paun", 2008, problema 1

[Claudiu Mindrila]

Fie \( a,b,c \) numere reale strict pozitive astfel incat \( a^2+b^2+c^2=1 \). Demonstrati ca:
\( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq 3+\frac{2(a^3+b^3+c^3)}{abc} \).

[Marius Mainea]

\( \sum{\frac{1}{a^2}}=\sum{\frac{a^2+b^2+c^2}{a^2}}=3+\sum{(\frac{b^2}{a^2}+\frac{c^2}{a^2})}=3+\sum{(\frac{a^2}{b^2}+\frac{a^2}{c^2)}\ge 3+\sum{2\frac{a}{b}\cdot \frac{a}{c}} \)