tag:blogger.com,1999:blog-444408592833092365.post3654953828123178534..comments2023-06-29T04:02:58.043-06:00Comments on México rumbo a la IMO: Problema del miércolesDavid (sirio11)http://www.blogger.com/profile/13765612869477578855noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-444408592833092365.post-45989925511899780942022-03-04T07:24:24.056-06:002022-03-04T07:24:24.056-06:00The Casino Floor, Las Vegas, NV (2021 Update)
The ...The Casino Floor, Las Vegas, NV (2021 Update)<br />The Casino Floor <a href="https://drmcd.com/%ec%84%b1%eb%82%a8%ec%97%90-%eb%8c%80%ed%95%9c-%ec%9a%b0%ec%88%98%ed%95%9c%ec%b6%9c%ec%9e%a5%ec%83%b5%eb%a6%ac%eb%b7%b0.html" rel="nofollow">성남 출장안마</a> is <a href="https://www.mapyro.com/%ea%b0%95%eb%a6%89%ec%b6%9c%ec%9e%a5%ec%95%88%eb%a7%88%ea%b0%80%ec%9d%b4%eb%93%9c.html" rel="nofollow">강릉 출장안마</a> located <a href="https://www.jtmhub.com/%ec%9b%90%ec%a3%bc%ea%b0%80%ec%9e%a5-%ec%9d%b8%ea%b8%b0-%ec%9e%88%eb%8a%94%ec%b6%9c%ec%9e%a5%eb%a7%88%ec%82%ac%ec%a7%80%ea%b0%80%ea%b2%8c.html" rel="nofollow">원주 출장안마</a> in the <a href="https://drmcd.com/%ec%84%b1%eb%82%a8%ec%97%90-%eb%8c%80%ed%95%9c-%ec%9a%b0%ec%88%98%ed%95%9c%ec%b6%9c%ec%9e%a5%ec%95%88%eb%a7%88%eb%a6%ac%eb%b7%b0.html" rel="nofollow">성남 출장안마</a> middle of the Strip. The <a href="https://www.jtmhub.com/%ec%98%81%ec%b2%9c%ec%b5%9c%ea%b3%a0%ec%8b%9c%ec%84%a4%ec%b6%9c%ec%9e%a5%eb%a7%88%ec%82%ac%ec%a7%80.html" rel="nofollow">영천 출장마사지</a> casino floor features 12 slot machines including the newest three-disc and five-zaidebabbithttps://www.blogger.com/profile/13918049632725268673noreply@blogger.comtag:blogger.com,1999:blog-444408592833092365.post-27600090693141261182017-06-07T14:21:16.827-05:002017-06-07T14:21:16.827-05:00Sea $p = 2 + x, q = 2 + y, r = 2 + z, s = 2 + w$, ...Sea $p = 2 + x, q = 2 + y, r = 2 + z, s = 2 + w$, y $x > y > z > w$ sin pérdida de generalidad, entonces las condiciones se transforman en $x + y + z + w = 1$ y $x^2 + y^2 + z^2 + w^2 = 1$. Ahora, buscamos probar que <br /><br />$$(2 + x)(2 + y) - (2 + z)(2 + w) \geq 2$$<br />$$\iff$$<br />$$2(x + y - z - w) + xy - wz \geq 2$$<br />$$\iff$$<br />$$8(x + y) + 2(xy - wz) \geq 8$$<br /><br />Donde usamos que $x + y + z + w = 1$. Ahora, tenemos que<br /><br />$$(x + y)^2 + (w - z)^2 = x^2 + y^2 + z^2 + w^2 + 2(xy - wz) = 1 + 2(xy - wz)$$<br /><br />Y por lo tanto la desigualdad deseada es equivalente a<br /><br />$$8(x + y) + (x + y)^2 + (z - w)^2 \geq 9$$<br /><br />Demostramos ahora que $x + y \geq 1$ con lo cual esta desigualdad es evidente. Suponemos lo contrario, entonces $z + w \textgreater 0$. Expandiendo $(x + y + z + w)^2$ encontramos que<br /><br />$$(x + y)(w + z) + xy + wz = xy + xz + xw + yz + yw + zw = 0$$<br /><br />Y por lo tanto es imposible que $x, y, z, w$ sean todos positivos. Luego $w \leq 0$ y por lo tanto $z$ es positivo. Entonces debemos tener que $xy + wz \textless 0$, y ya que $|z| < |y|$ esto implica que $|w| > |x|$. Ya que $w$ es negativo y $x + y + z + w = 1$ esto implica que $y + z \textgreater 1$, y entonces $y^2 + z^2 \textgreater \frac{1}{2}$. Además $x \textgreater y \textgreater \frac{1}{2}$ y $w \textless -x < -\frac{1}{2}$. Esto junto implica que<br /><br />$$x^2 + y^2 + z^2 + w^2 \textgreater 1$$<br /><br />Contradicción. Luego $x + y \geq 1$ y terminamos.Arielhttps://www.blogger.com/profile/01448135590361074114noreply@blogger.com