mateforum.ro

a) Descompuneti in factori expresia \( 3a^{4}-2a^{3}b-2a^{2}b^{2}-2ab^{3}+3b^{4} \).

b) Aratati ca pentru orice numere reale \( a,b\in\mathbb{R} \) are loc inegalitatea \( \frac{a^{4}+b^{4}}{2}\ge\left(\frac{a+b}{2}\right)^{2}\cdot\frac{a^{2}+b^{2}}{2} \).

Concursul "TMMATE", 2009

b) Folosim rezultatul cunoscut:
\( \frac{a^n+b^n}{2} \ge (\frac{a+b}{2})^n \)

Avem \( \frac{a^4+b^4}{2}\ge (\frac{a^2+b^2}{2})^2=\frac{a^2+b^2}{2}\cdot \frac{a^2+b^2}{2} \) si aplicand inca o data avem \( \frac{a^4+b^4}{2}\ge (\frac{a+b}{2})^2 \cdot \frac{a^2+b^2}{2} \).

a) \( P = 3(a^2-b^2)^2 + 6a^2b^2 - 2a^3b - 2a^2b^2 - 2ab^3 \) \( =3(a-b)^2(a+b)^2 -2ab(a-b)^2 \)\( =(a-b)^2(3(a+b)^2 - 2ab) \) s.a.m.d.