tag:blogger.com,1999:blog-2400513859305780710.post5600195178989867638..comments2023-04-04T07:53:53.789-07:00Comments on Mathematical Food For Thought: Hey Now. Topic: Inequalities. Level: AIME.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2400513859305780710.post-43520830283793959132007-03-26T17:34:36.000-07:002007-03-26T17:34:36.000-07:00Hm...$ \displaystyle \sum_{sym} a^2b+2abc\ge 2+2(a...Hm...<br><br>$ \displaystyle \sum_{sym} a^2b+2abc\ge 2+2(a+b+c)$<br><br>So maybe we can prove $a^2(b+c)\ge 2a$.<br><br>Or $\frac{1}{b}+\frac{1}{c}\ge 2$<br><br>This is correct for $a>1$ because of $\frac{1}{b}+\frac{1}{c}\ge 2\sqrt{a}\ge 2$ by AM-GM.ch1n353ch3s54a1lnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-62368301141339926352007-03-27T10:31:16.000-07:002007-03-27T10:31:16.000-07:00It's not always true though ;). Try $ a = 1/10...It's not always true though ;). Try $ a = 1/10, b = 10, c = 1$.paladin8noreply@blogger.com