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Let \(x,y,z \ge 0\) where \(x+y+z=1\). Prove that

\[ \large{x^3+y^3+z^3+6xyz \ge \dfrac{1}{4} }.\]

Note by Lakshya Sinha 1 year, 11 months ago

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\(\frac{1}{4}=\frac{(x+y+z)^3}{4}\)

\(x^3+y^3+z^3+6xyz\geq\frac{(x+y+z)^3}{4}\)

\(4x^3+4y^3+4z^3+24xyz\geq x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2+6xyz\)

\(3x^3+3y^3+3z^3+18xyz\geq3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2\)

\(x^3+y^3+z^3+6xyz\geq x^2y+xy^2+y^2z+yz^2+z^2x+zx^2\)

Which is true by Schur's. Equality when two of \(x, y, z\) equal \(\frac{1}{2}\) and the third equals \(0\). Note that the original inequality would still hold if it were \(x^3+y^3+z^3+3.75xyz\) instead of \(x^3+y^3+z^3+6xyz\)

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Wrote very nicely+1

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## Comments

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TopNewest\(\frac{1}{4}=\frac{(x+y+z)^3}{4}\)

\(x^3+y^3+z^3+6xyz\geq\frac{(x+y+z)^3}{4}\)

\(4x^3+4y^3+4z^3+24xyz\geq x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2+6xyz\)

\(3x^3+3y^3+3z^3+18xyz\geq3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2\)

\(x^3+y^3+z^3+6xyz\geq x^2y+xy^2+y^2z+yz^2+z^2x+zx^2\)

Which is true by Schur's. Equality when two of \(x, y, z\) equal \(\frac{1}{2}\) and the third equals \(0\). Note that the original inequality would still hold if it were \(x^3+y^3+z^3+3.75xyz\) instead of \(x^3+y^3+z^3+6xyz\)

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Wrote very nicely+1

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