Prove That A^2+B^2+C^2

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This is my proof. I"m not sure if it is true but perhaps someone would tell me yes or (no & why).

Since $(a^2+b^2), (b^2+c^2), (c^2+a^2)$ are greater than or equal khổng lồ zero, then $(a^2+b^2)(b^2+c^2)(c^2+a^2) geq 0$. I"ll spare you the distribution and say this, $(a^2+b^2)(b^2+c^2)(c^2+a^2) + 8a^2b^2c^2 geq 8a^2b^2c^2$

Done.

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$egingroup$ You actually need khổng lồ show $(a^2+b^2)(b^2+c^2)(c^2+a^2)geq 8a^2b^2c^2$... I bởi not understvà why would $(a^2+b^2)(b^2+c^2)(c^2+a^2) + 8a^2b^2c^2 geq 8a^2b^2c^2$ imply $(a^2+b^2)(b^2+c^2)(c^2+a^2)geq 8a^2b^2c^2$ $endgroup$–user87543 Dec 17 "13 at 2:21
AndresCaicebởi vì hint.... you need to apply that thrice.... Good luck! $endgroup$–user87543 Dec 17 "13 at 2:24
rewriting what andres said : $(a^2+b^2)geq2ab,(b^2+c^2)geq2bc,(c^2+a^2)geq2ca$.

therefore $(a^2+b^2)(b^2+c^2)(c^2+a^2) geq (2ab)(2bc)(2ca) = 8a^2 b^2 c^2$


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$egingroup$ I have down voted this... It would not be a better idea to lớn write complete solution if some other person has given hints just before ten minutes... you could have sầu given some time... $endgroup$–user87543 Dec 17 "13 at 2:33
$egingroup$ seeing a solution is not that bad... the person who asked the question should see the solution by trying a bit hard with given hint... any way it would be their personal choice... $endgroup$–user87543 Dec 17 "13 at 2:41

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