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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 AM-GM, $$\frac{ab+bc+ac}{3}\ge\sqrt[3]{a^2b^2c^2}$$ The result follows. Thanks from MIKI14
 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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