a) Sa se arate ca \( x^2+y^2+1\ge \4sqrt{xy} \) pentru orice x,y pozitive.
b) Determinati numerele pozitive \( a_1,a_2,..,a_{2005} \) stiind ca \( \frac{1}{a_1^2+a_2^2+2}+\frac{1}{a_2^2+a_3^2+2}+...+\frac{1}{a_{2005}^2+a_1^2+2}=\frac{1}{4}\(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_{2005}\) \)
D. Piciu, Concursul Gh.Titeica
Consecinta unei inegalitati
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Marius Mainea
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Consecinta unei inegalitati
Last edited by Marius Mainea on Sat Apr 18, 2009 5:32 pm, edited 1 time in total.
a) Din inegalitatea mediilor \( \frac{x+y}{2}\ge \sqrt{xy} \).
Trebuie sa demonstram ca \( x^2+y^2+1\ge \frac{x+y}{2} \), adica \( 2x^2+2y^2+2\ge x+y \), \( 2x^2-x+2y^2-y+2\ge 0 \).
Dar \( (\sqrt{2}x-1)^2+(\sqrt{2}y-1)^2\ge 0 \)
\( 2x^2-2\sqrt{2}x+1+2y^2-2\sqrt{2}y+1\ge 0 \)
Trebuie sa demonstram ca \( 2x^2+2y^2+2-(x+y)\ge sx^2+2y^2+2-2\sqrt{2}(x+y) \)
\( -(x+y)\ge -2\sqrt{2}(x+y) \)
\( x+y\le 2\sqrt{2}(x+y) \)
\( 1\le 2sqrt{2} \), adevarat
Trebuie sa demonstram ca \( x^2+y^2+1\ge \frac{x+y}{2} \), adica \( 2x^2+2y^2+2\ge x+y \), \( 2x^2-x+2y^2-y+2\ge 0 \).
Dar \( (\sqrt{2}x-1)^2+(\sqrt{2}y-1)^2\ge 0 \)
\( 2x^2-2\sqrt{2}x+1+2y^2-2\sqrt{2}y+1\ge 0 \)
Trebuie sa demonstram ca \( 2x^2+2y^2+2-(x+y)\ge sx^2+2y^2+2-2\sqrt{2}(x+y) \)
\( -(x+y)\ge -2\sqrt{2}(x+y) \)
\( x+y\le 2\sqrt{2}(x+y) \)
\( 1\le 2sqrt{2} \), adevarat