Prove x6+y6+z6+3x2y2z2≥2(x3y3+y3z3+z3x3)x^6 + y^6 + z^6 + 3x^2y^2z^2 \geq 2(x^3y^3 + y^3z^3 + z^3x^3) [on hold]

For x,y,z>0x,y,z>0, prove
x6+y6+z6+3x2y2z2≥2(x3y3+y3z3+z3x3)
x^6 + y^6 + z^6 + 3x^2y^2z^2 \geq 2(x^3y^3 + y^3z^3 + z^3x^3)

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1 Answer
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By Schur and Muirhead
∑cyc(x6+x2y2z2)≥∑cyc(x4y2+x4z2)≥2∑cycx3y3\sum\limits_{cyc}(x^6+x^2y^2z^2)\geq\sum\limits_{cyc}(x^4y^2+x^4z^2)\geq2\sum\limits_{cyc}x^3y^3
Done!

  

 

Tnx, Schur was the crucial step. This can’t be done only with Muirhead.
– Maki
2 days ago

  

 

@Mak Of course! ∑cyc(x4y2+x4z2−2x3y3)=∑cycx2y2(x−y)2≥0\sum\limits_{cyc}(x^4y^2+x^4z^2-2x^3y^3)=\sum\limits_{cyc}x^2y^2(x-y)^2\geq0
– Michael Rozenberg
yesterday