tag:blogger.com,1999:blog-2400513859305780710.post7130689433453000314..comments2023-04-04T07:53:53.789-07:00Comments on Mathematical Food For Thought: Don't Mess With The Juggernaut. Topic: Inequalities. Level: Olympiad.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-2400513859305780710.post-24929880407541228212006-05-30T11:01:23.000-07:002006-05-30T11:01:23.000-07:00(Wow, when I first saw that problem I thought it w...(Wow, when I first saw that problem I thought it was hard.)<br><br>We can factor!<br><br>(2x - \lfloor x-1 \rfloor)(x - \lfloor x-1 \rfloor) \le 0<br><br>Let u = \lfloor x-1 \rfloor and let d = fpart(x-1). Then<br><br>(2(u + d + 1) - u)(u + d + 1 - u) \le 0<br>(u + 2d + 1)(d + 1) \le 0<br><br>The second factor is always positive, so the first must be negative. The first solution comes at x = -1.<br><br>For -2 \le x < -1 we have u = -2 and hence we want (2d - 1) \le 0. This requires d \le 1/2, and o ur second solution set comes at -2 \le x \le -1.5.<br><br>For x < -2 we have u = -3 and the first factor will always be negative.<br><br>QED.QCnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-89208526740139060622006-05-30T11:06:02.000-07:002006-05-30T11:06:02.000-07:00Oh snap the first factor should be (u + 2d + 2). ...Oh snap the first factor should be (u + 2d + 2). And for -2 \le x < -1 we have u = -3, so my mistakes cancel out and the solutions are the same XD<br><br>Alternately, \lfloor x-1 \rfloor = \lfloor x \rfloor - 1, which simplifies the problem a bit and makes it harder to make dumb mistakes.QCnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-5091020912531512902006-05-30T11:49:24.000-07:002006-05-30T11:49:24.000-07:00Yeah, another way to do it is to set a = x, b = [[...Yeah, another way to do it is to set a = x, b = [[x-1]] and factor that. It was on AoPS, but they missed the negative solutions.paladin8noreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-60581920539749742342010-08-09T18:51:44.000-07:002010-08-09T18:51:44.000-07:00[...] have the minimum result generalized here. Th...[...] have the minimum result generalized here. The proof is the same [...]Inequality a^3+b^3+c^3+8abc | MathFax.comhttp://mathfax.com/inequality-a3b3c38abc/noreply@blogger.com