Solution
First we show that for m \geq 8 it's true. By Cauchy: \frac{x}{x^2 + myz} + \frac{y}{y^2 + mxz} + \frac{z}{z^2 + mxy} \geq \frac{(x+y+z)^2}{x^3 + y^3 + z^3 + 3m.xyz} Then let A = x^2 + y^2 + z^2,B = xy+yz+zx. Because x^3+y^3+z^3 = (x+y+z)(A -B)+3xyz, = \frac{A+2B}{(x+y+z)(A-B)+3(m+1)xyz} \geq \frac{9}{(m+1)(x+y+z)} The last inequality is equivalent to (because A \geq B we may multiply by the denominators without changing the sign): (A+2B)(x+y+z)(m+1) \geq 9(x+y+z)(A-B) + 27(m+1)(x+y+z) \iff (x+y+z)[(m-8)A + (2m+11)B] \geq 27(m+1)xyz Indeed by AM-GM: x+y+z \geq 3 \sqrt[3]{xyz} A \geq 3 \sqrt[3]{x^2y^2z^2} B \geq 3 \sqrt[3]{x^2y^2z^2} So: (x+y+z)[(m-8)A + (2m+11)B] \geq 3 \sqrt[3]{xyz} [3(m-8) \sqrt[3]{x^2y^2z^2} + 3(2m+11)\sqrt[3]{x^2y^2z^2} = 27(m+1)xyz Now to show necessity, plug in z = t, x=y=1, then the left hand side is : \frac{2}{1+mt} + \frac{t}{t^2+m} \geq \frac{9}{(m+1)(2+t)} (2(t^2+m)+t(mt+1))(2+t)(m+1) \geq 9(t^2+m)(mt+1) which has to be true for all t \geq 0. However, the coefficient of t^3 on the LHS is m(m+1) and on the RHS is 9m. Because this inequality has to be true for all t no matter how big, then the coefficient on the LHS has to be greater than or equal to that of the RHS, otherwise we can choose t large enough to violate that expression. m(m+1) \geq 9m which means m \geq 8.
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