mateforum.ro

Daca \( a,b,c\geq 0 \) si \( a+b+c=1 \) aratati ca:

\( \frac{a}{\sqrt{b^2+3c}}+\frac{b}{\sqrt{c^2+3a}}+\frac{c}{\sqrt{a^2+3b}}\geq \frac{1}{\sqrt{1+3abc}} \)

Din inegalitatea Holder avem:

\( \left(\sum_{cyc}{\frac{a}{\sqrt{b^{2}+3c}}}\right)^{2}\left(\sum_{cyc}{a(b^{2}+3c)}\right) \geq 1 \).

Deci ne mai ramane sa aratam ca \( 1+3abc \geq \sum_{cyc}{ab^{2}}+3\sum_{cyc}{ab} \).

Care dupa omogenizare (\( a+b+c=1 \)), se reduce la \( \sum_{cyc}{a^{3}} \geq \sum_{cyc}{ab^{2}} \), care este adevarata.

Ce zici de urmatoarea :
Daca \( a,b,c\geq 0 \) si \( a+b+c=1 \)
\( \sum \frac{b}{\sqrt{a^2+3b}}\leq \frac{1}{\sqrt{1+3abc}} \)