Inegalitate

[alex2008]

Sa se arate ca daca \( a, b, c > 0 \) si \( abc=\frac{1}{3} \) , atunci are loc inegalitatea :
\( \sum_{cyc} \frac{3a^5}{3a^5+2bc} \ge 1 \)



Costel Anghel , ,,Nicolae Coculescu,, , Ziua 1 , 30 noiembrie 2007 , Slatina

[maxim bogdan]

Avem: \( \sum_{cyc}\frac{3a^5}{3a^5+2bc}=\sum_{cyc}\frac{a^6}{a^6+\frac{2}{9}}\geq \frac{(a^3+b^3+c^3)^2}{a^6+b^6+c^6+\frac{2}{3}}\geq 1. \) (am aplicat inegalitatea Cauchy-Schwarz)\( \Longleftrightarrow \)

\( \Longleftrightarrow\sum_{cyc}a^3b^3\geq \frac{1}{3}. \)

Din inegalitatea mediilor avem: \( a^3b^3+b^3c^3+c^3a^3\geq 3\sqrt[3]{a^6b^6c^6}=\frac{1}{3}, \) care este chiar inegalitatea dorita.

[DrAGos Calinescu]

Sau \( \sum_{cyc}\frac{3a^5}{3a^5+2bc}=\sum_{cyc}\frac{a^6}{a^6+\frac{2}{3}}=\sum_{cyc}\frac{9a^6}{9a^6+2} \)
Fie \( a^6=x, b^6=y, c^6=z\Rightarrow xyz=\frac{1}{3^6} \)

Prin aducere la acelasi numitor se obtine \( \frac {324(xy+yz+zx)+36(x+y+z)+3}{162(xy+yz+zx)+36(x+y+z)+9}\ge 1\Leftrightarrow xy+yz+zx\ge \frac{1}{27}\ \)
\( \frac{xy+yz+zx}{3}\ge\sqrt[3]{\frac{1}{3^{12}}}=\frac{1}{3^4}\Rightarrow xy+yz+zx\ge\frac{1}{27} \)