
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
August 23rd, 2018, 02:44 AM  #1 
Newbie Joined: Dec 2017 From: Spain Posts: 18 Thanks: 1  Easy inequality
How can I prove: (ab + bc + ca)^3 ≥ 27[(abc)^2] 
August 23rd, 2018, 02:48 AM  #2 
Newbie Joined: Dec 2017 From: Spain Posts: 18 Thanks: 1 
This is equivalent to proving: (ab+bc+ca)/3 ≥ √[(abc)^2] which somehow reminds me of the arithmetic and geometric mean inequality, although I don't really know how to apply it... Thanks in advance 
August 23rd, 2018, 05:08 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
From the AMGM, $$\frac{ab+bc+ac}{3}\ge\sqrt[3]{a^2b^2c^2}$$ The result follows. 
August 23rd, 2018, 06:19 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,882 Thanks: 1835 
The inequality doesn't hold for (a, b, c) = (1, 1, 1).


Tags 
easy, inequality 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Triangle Inequality: Prove Absolute Value Inequality  StillAlive  Calculus  5  September 3rd, 2016 12:45 AM 
Easy inequality question  jyrdo  Algebra  6  April 3rd, 2014 09:55 PM 
Help Easy  Algebra  5  January 27th, 2011 11:56 AM  
an area inequality(may be very easy)  earth  Algebra  0  September 8th, 2010 12:15 AM 
an area inequality(may be very easy)  earth  Math Events  0  December 31st, 1969 04:00 PM 