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ʹء¡ÑºÍÊÁ¡ÒÃ
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Show that for arbitrary $a.b.c \in \mathbb{R^{+}}$, $$\frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}} \geq \frac{a+b+c}{3}$$ ¢Íº¤Ø³ÁÒ¡¤ÃѺ |
#2
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from
$\frac{a^3}{a^2+ab+b^2}\geq \frac{2a-b}{3}$ the rest is easy ;-)
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Rose_joker @Thailand Serendipity |
#3
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âË ¤Ø³ Rose Âѧà¢éÒÁÒàÅè¹ã¹áÁ·à«¹àµÍÃì´éÇÂ
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#4
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àÇÅÒàÃÒ·Ó·ÓÍÂèÒ§¹ÕéËÃͤѺ
â´ÂäÁèàÊչѷÑèÇ仡Ó˹´ãËé $a \geq b \geq c$ áÅéÇàÃÒ¡çÍéÒ§ $\frac{a^3}{a^2+ab+b^2}\geq \frac{2a-b}{3}$, $\frac{a^3}{b^2+bc+c^2}\geq \frac{2b-c}{3}$, $\frac{c^3}{c^2+ca+a^2}\geq \frac{2c-a}{3}$ áÅéǨѺÁҺǡ¡Ñ¹ËÃ×Íà»ÅèÒ¤ÃѺ |
#5
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ÍéÒ§ÍÔ§:
ÇÔ¸Õà¢Õ¹¡çáÅéÇáµèÊäµÅ줹à¢Õ¹¤ÃѺ ¡çÍÒ¨¨Ðà¢Õ¹Ẻ¹Õé ¨Ò¡ AM-GM ä´éÇèÒ $a^3+b^3\geq a^2b+ab^2$_____(*) $3a^3\geq 2a^3-a^2b+2a^2b-ab^2+2ab^2-b^3$ $3a^3\geq (2a-b)(a^2+ab+b^2)$ $\displaystyle\therefore\frac{a^3}{a^2+ab+b^2}\geq\frac{2a-b}{3}$ 㹷ӹͧà´ÕÂǡѹ ä´éÇèÒ $\displaystyle\frac{b^3}{b^2+bc+c^2}\geq \frac{2b-c}{3}$, $\displaystyle\frac{c^3}{c^2+ca+a^2}\geq \frac{2c-a}{3}$ ´Ñ§¹Ñé¹ $\displaystyle\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\geq\frac{a+b+c}{3}$ ËÃ×ÍÇèÒ¨Ðà¢Õ¹Ẻ¹Õé ⨷ÂìÊÁÁÙšѺ $\displaystyle\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\geq\frac{a+b+c}{3}=\sum_{cyc}\frac{2a-b}{3}$ ´Ñ§¹Ñé¹ à»ç¹¡ÒÃà¾Õ§¾Í·Õè¨ÐáÊ´§ÇèÒ $\displaystyle\frac{a^3}{a^2+ab+b^2}\geq\frac{2a-b}{3}$ ¨Ò¡ $\displaystyle\frac{a^3}{a^2+ab+b^2}\geq\frac{2a-b}{3}$ $\Leftrightarrow 3a^3\geq 2a^3-a^2b+2a^2b-ab^2+2ab^2-b^3$ $\Leftrightarrow a^3+b^3\geq a^2b+ab^2$ «Öè§àËç¹ä´éªÑ´ÇèÒà»ç¹¨ÃÔ§¨Ò¡ AM-GM_____(*) ».Å. áµè¶éÒ¨ÐãËéÃÑ´¡ØÁ¡ÇèÒ¹Õé ¡çÍÒ¨¨Ðà¢Õ¹¾ÔÊÙ¨¹ìÇèÒ·ÓäÁµÃ§ (*) Áѹ¶Ö§à»ç¹¨ÃÔ§¡çä´é¤ÃѺ 27 àÁÉÒ¹ 2009 14:34 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ owlpenguin à˵ؼÅ: á¡é¾ÔÁ¾ì¼Ô´ |
#6
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¢Íº¤Ø³¼ÙéªèÇÂ͹Øà¤ÃÒÐËì¼Á·Ø¡¤¹ÁÒ¡æ¤ÃѺ
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#7
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ÍéÒ§ÍÔ§:
¡è͹Í×è¹ÊѧࡵÇèÒ $\dfrac{a^2-ab+b^2}{a^2+ab+b^2}\geq\dfrac{1}{3}$ «Öè§ÍÊÁ¡ÒÃÊÁÁÙšѺ $(a-b)^2\geq 0$ áÅÐÊѧࡵÇèÒ $\dfrac{a^3-b^3}{a^2+ab+b^2}+\dfrac{b^3-c^3}{b^2+bc+c^2}+\dfrac{c^3-a^3}{c^2+ca+a^2}=(a-b)+(b-c)+(c-a)=0$ ´Ñ§¹Ñé¹ $\dfrac{a^{3}}{a^{2}+ab+b^{2}}+\dfrac{b^{3}}{b^{2}+bc+c^{2}}+\dfrac{c^{3}}{c^{2}+ca+a^{2}}=\dfrac{b^{3}}{a^{2}+ab+b^{2}}+\dfrac{ c^{3}}{b^{2}+bc+c^{2}}+\dfrac{a^{3}}{c^{2}+ca+a^{2}}$ ¨Ö§ä´é $\dfrac{a^{3}}{a^{2}+ab+b^{2}}+\dfrac{b^{3}}{b^{2}+bc+c^{2}}+\dfrac{c^{3}}{c^{2}+ca+a^{2}}=\dfrac{1}{2}\Big(\dfrac{a^{3}+b^3}{a^ {2}+ab+b^{2}}+\dfrac{b^{3}+c^3}{b^{2}+bc+c^{2}}+\dfrac{c^{3}+a^3}{c^{2}+ca+a^{2}}\Big)$ $~~~~~=\dfrac{1}{2}\Big[(a+b)\Big(\dfrac{a^2-ab+b^2}{a^2+ab+b^2}\Big)+(b+c)\Big(\dfrac{b^2-bc+c^2}{b^2+bc+c^2}\Big)+(c+a)\Big(\dfrac{c^2-ca+a^2}{c^2+ca+a^2}\Big)\Big]$ $~~~~~\geq\dfrac{1}{2}\Big(\dfrac{a+b}{3}+\dfrac{b+c}{3}+\dfrac{c+a}{3}\Big)$ $~~~~~=\dfrac{a+b+c}{3}$
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site:mathcenter.net ¤Ó¤é¹ |
#8
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¢Íº¤Ø³¾Õè˹ØèÂÍÕ¡¤¹¹èФÃѺ
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#9
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ÍéÒ§ÍÔ§:
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#10
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¨Ò¡â¨·Âì â´ÂÍÊÁ¡ÒÃ⤪Õä´éÇèÒ $L.H.S = \sum_{cyc}\frac{a^4}{a^3+a^2b+ab^2}\geqslant \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+c^2a+ca^2} =\frac{a^2+b^2+c^2}{a+b+c}$
«Öè§â´ÂÍÊÁ¡Òà Power mean ä´éÇèÒ $\frac{a^2+b^2+c^2}{a+b+c} \geqslant \frac{a+b+c}{3}$ µÒÁµéͧ¡Òà 09 ¾ÄÉÀÒ¤Á 2009 22:33 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ S!xTo12Y à˵ؼÅ: á¡é Latex |
#11
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à¾Ôè§à¨ÍÍÊÁ¡Ò÷Õè sharp ¡ÇèÒ¤ÃѺ ÊÓËÃѺÇÔ¸Õ¤Ô´¡ç¤§ÃÙé¡Ñ¹áÅéÇÅèÐ
$a,b,c>0$ $$\frac{a^{3}}{a^{2}+ab+b^{2}}+\frac{b^{3}}{b^{2}+bc+c^{2}}+\frac{c^{3}}{c^{2}+ca+a^{2}} \geq \frac{2}{3}\Big(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\Big) \geq \frac{a+b+c}{3}$$
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site:mathcenter.net ¤Ó¤é¹ |
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