O noua inegalitate
Posted: Fri Jul 11, 2008 8:06 pm
Daca \( a,b,c>0 \) si \( a^2+b^2+c^2=1 \), atunci
\( \frac{a^5+b^5}{ab(a+b)}+\frac{b^5+c^5}{bc(b+c)}+\frac{c^5+a^5}{ca(c+a} \geq 3(ab+bc+ca)-2 \).
\( \frac{a^5+b^5}{ab(a+b)}+\frac{b^5+c^5}{bc(b+c)}+\frac{c^5+a^5}{ca(c+a} \geq 3(ab+bc+ca)-2 \).