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@Sualeh Asif
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I found this amazingly elegant proof on AoPS: (a2+b2+c2)2−3(a3b+b3c+c3a)=21((a2−2ab+bc−c2+ca)2+(b2−2bc+ca−a2+ab)2+(c2−2ca+ab−b2+bc)2)≥0
@Abdur Rehman Zahid
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It isnt very difficult that way. Try yourself expanding the whole inequality,without any denominators. Then work with AM-GM and Shur. (And post the expanded stuff)
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Here is a short solution I found:
cyc∑xz3+x3zx4≥2(xz3+x3y+y3z)(x2+y2+z2)2≥23
Hence it suffices to prove that
(x2+y2+z2)2≥3(xz3+x3y+y3z)
Which is the famous Vasc's inequality
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Well, proving the Vasc Inequality might be slightly more difficult.
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To be exact Tremendously difficult
It does not yeild to most classical inequalities!
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AoPS:
I found this amazingly elegant proof on(a2+b2+c2)2−3(a3b+b3c+c3a)=21((a2−2ab+bc−c2+ca)2+(b2−2bc+ca−a2+ab)2+(c2−2ca+ab−b2+bc)2)≥0
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Yep! Titu and Vasc killed this... =D
@MS HT @Sharky Kesa @ZK LIn
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You really want my long, expanded solution?
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Yes :P
Yes please.
I still haven't seen your long and expanded solution Sharky :P
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@Gurīdo Cuong