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August 23rd, 2018, 02:44 AM   #1
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Easy inequality

How can I prove:

(ab + bc + ca)^3 ≥ 27[(abc)^2]
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August 23rd, 2018, 02:48 AM   #2
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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
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August 23rd, 2018, 05:08 AM   #3
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From the AM-GM,

$$\frac{ab+bc+ac}{3}\ge\sqrt[3]{a^2b^2c^2}$$

The result follows.
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August 23rd, 2018, 06:19 AM   #4
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The inequality doesn't hold for (a, b, c) = (1, 1, -1).
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