Inegalitate

[alex2008]

Daca \( a,b,c>0 \) astfel incat \( abc=1 \) , sa se arate ca :

\( \sum_{cyc}\sqrt{\frac{b+c}{a^3+1}}\ge \frac{9}{a+b+c} \).

Concursul Arhimede

[Marius Mainea]

LHS=\( \sum {\sqrt{\frac{b+c}{a(a^2+bc)}}}\ge\sum {\frac{2}{a+\frac{a^2+bc}{b+c}}}=2\sum {\frac{b+c}{(a+b)(a+c)}}\ge \)RHS