Bihor 2008

[alex2008]

Daca \( a , b , c >0 \) si \( abc=1 \) , aratati ca :
\( E=\frac{1}{a^2(b+c)}+\frac{1}{b^2(a+c)}+\frac{1}{c^2(a+b)}\ge\frac{3}{2} \)

[Marius Mainea]

Folosind inegalitatea CBS avem :
\( LHS=\sum{\frac{b^2c^2}{b+c}}\ge \frac{(bc+ca+ab)^2}{b+c+c+a+a+b}\ge\frac{3abc(a+b+c)}{2(a+b+c)}=\frac{3}{2} \)