Find all m for inequality

Find all real number $m$ such that, for all positive real numbers $x,y,z$ the following is true: $$ \frac{x}{x^2 + myz} + \frac{y}{y^2 + mxz} + \frac{z}{z^2 + mxy} \geq \frac{9}{(m+1)(x+y+z)}$$

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 $m \geq 8$.