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Concursul "Teodor Topan" - problema 3
Posted: Mon Dec 03, 2007 2:21 pm
by maky
Fie \( a,b\in\left[o,\infty\right) \) astfel incat \( \begin{cases}
a\le1 \\ a+b\le10\end{cases} \). Aratati ca \( \sqrt{a}+\sqrt{b}\le4 \).
Vlaicu Liviu
Posted: Mon Dec 03, 2007 2:23 pm
by maky
philandrew wrote:
Din Cauchy,
\( (\sqrt{a} + \sqrt{b})^2 \) \( = \left( \sqrt{a} + \frac{\sqrt{b}}{3} + \frac{\sqrt{b}}{3} + \frac{\sqrt{b}}{3} \right)^2 \) \( \le 4 \left( a + \frac{b}{9} + \frac{b}{9} + \frac{b}{9} \right) \) \( = 4 \cdot \frac{2a + (a + b)}{3} \le 4 \cdot \frac{2 + 10}{3} = 16 \), deci
\( \sqrt{a} + \sqrt{b} \le 4 \), cu egalitate iff
\( a=1 \) si
\( b=9 \). De fapt si acest gen de inegalitati este clasic
Amatorii pot gasi un alt exercitiu de acest tip
aici.