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Inegalitati.
Posted: Sun Nov 30, 2008 2:31 pm
by Marius Mainea
Aratati ca:
a) \( u^2+v^2+uv\ge 0 , (\forall) u,v \in\mathbb{R} \)
b) \( x^2+y^2+xy-3x-3y+3\ge 0 , (\forall) x,y\in\mathbb{R} \)
c) \( \frac{a+b-1}{a^2+b^2+ab}+\frac{b+c-1}{b^2+c^2+bc}+\frac{c+a-1}{c^2+a^2+ca}\le 1 ,(\forall) a,b,c\in\mathbb{R} \)
V. Cornea, Dan St. Marinescu , Concursul ,,Nicolae Paun'',2003
Posted: Sun Nov 30, 2008 11:05 pm
by Claudiu Mindrila
a)
Daca \( u \leq v \), atunci:
\( u^2+v^2+uv \geq 0 \Longleftrightarrow (v-u)(u^2+v^2+uv) \geq 0 \Longleftrightarrow v^3-u^3 \geq 0 \Longleftrightarrow v^3 \geq u^3 \). Analog daca \( u>v \) rezulta \( u^3>v^3 \).
b)
Inegalitatea este echivalenta cu \( \left(x+ \frac{y-3}{2} \right) ^2+ \frac{3(y-1)^2}{4} \geq 0 \).
c) Deoarece \( \frac{a+b-1}{a^2+b^2+ab} \geq \frac{1}{3} \text(vezi b) ) \) rezulta ca \( \sum \frac{a+b-1}{a^2+b^2+ab} \geq 3 \cdot \frac{1}{3}=1 \).
Posted: Sun Nov 30, 2008 11:08 pm
by Marius Mainea
Claudiu Mindrila wrote:a)
Daca \( u \leq v \), atunci:
\( u^2+v^2+uv \geq 0 \Longleftrightarrow (v-u)(u^2+v^2+uv) \geq 0 \Longleftrightarrow v^3-u^3 \geq 0 \Longleftrightarrow v^3 \geq u^3 \). Analog daca \( u>v \) rezulta \( u^3>v^3 \).
b)
Inegalitatea este echivalenta cu \( \left(x+ \frac{y-3}{2} \right) ^2+ \frac{3(y-1)^2}{4} \geq 0 \).
c) Deoarece \( \frac{a+b-1}{a^2+b^2+ab} \leq \frac{1}{3} \text(vezi b) ) \) rezulta ca \( \sum \frac{a+b-1}{a^2+b^2+ab} \leq 3 \cdot \frac{1}{3}=1 \).