Inegalitate frumoasa, cam usurica totusi

[Claudiu Mindrila]

Sa se arate ca daca \( a,b,c \in (0, \infty) \), atunci:
\( \frac{3a^2-b^2-c^2}{bc}+\frac{3b^2-c^2-a^2}{ca}+\frac{3c^2-a^2-b^2}{ab} \geq 3. \)
Gheorghe Stoica

[Marius Mainea]

Prin aducere la acelasi numitor si desfacerea parantezelor obtinem:

\( \sum{(3a^2-b^2-c^2)a}\ge 3abc \) sau

\( 3\sum{a^3}\ge 3abc+\sum {a(b^2+c^2)} \) care este adevarata deoarece

\( a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ge 0 \) si

\( 2\sum{a^3}=\sum{(a^3+b^3)}=\sum{(a+b)(a^2-ab+b^2)}\ge\sum{(a+b)ab=\sum{a(b^2+c^2)} \)