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 May 11th, 2015, 11:44 PM #1 Newbie   Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0 Inequality problem Hi guys, here is a problem I'm stuck on: If $a$, $b$ and $c$ are positive reals, prove that $\sqrt{abc}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq2 \sqrt{3}$ I found out if $a=b=c$, then the LHS equals $a+\dfrac{3}{a}$. This is always $\geq2 \sqrt{3}$. But how to show the minimum is at $a=b=c$? Thanks in advance for any solutions. May 12th, 2015, 02:22 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 By the AM-GM inequality, $\dfrac1a + \dfrac1b + \dfrac1c \geqslant \dfrac{3}{\sqrt{abc}}$. By the AM-GM inequality, $\sqrt{abc} + \dfrac{3}{\sqrt{abc}} \geqslant 2\sqrt3$. Thanks from Olinguito May 12th, 2015, 02:36 AM #3 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Another method – start with: $\displaystyle \left[(abc)^{1/6}-\sqrt{\frac3a}\right]^2+\left[(abc)^{1/6}-\sqrt{\frac3b}\right]^2+\left[(abc)^{1/6}-\sqrt{\frac3c}\right]^2\ \geqslant\ 0$Expanding and rearranging: $\displaystyle 3\sqrt{abc} + 3\left(\frac1a+\frac1b+\frac1c\right)\ \geqslant\ 2\sqrt3(abc)^{1/6}\left(\frac1{\sqrt a}+\frac1{\sqrt b}+\frac1{\sqrt b}\right)$ Now apply AM–GM to $\frac1{\sqrt a}+\frac1{\sqrt b}+\frac1{\sqrt b}\geqslant\frac3{(abc)^{1/6}}$. May 12th, 2015, 02:56 AM #4 Newbie   Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0 Thanks for both of your solutions, but I think skipjack's solution is neater and simpler. Tags inequality, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post itiffanysphone Algebra 6 December 10th, 2014 08:13 AM wawar05 Algebra 1 November 2nd, 2012 04:39 AM Havoc Number Theory 1 October 25th, 2010 06:32 PM knowledgegain Algebra 1 May 6th, 2009 01:37 PM kokusaka Algebra 0 September 30th, 2008 07:45 AM

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