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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: 7,912 Thanks: 1110 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: 20,285 Thanks: 1967 
The inequality doesn't hold for (a, b, c) = (1, 1, 1).


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