# 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.

Bạn đang xem: Prove That A^2+B^2+C^2  \$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\$ \$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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