crazyfehmy wrote: Prove that \[ x^4y+y^4z+z^4x+xyz(x^3+y^3+z^3) \geq (x+y+z)(3xyz-1) \]for all positive real numbers $x, y, z$. By AM-GM, Rearrangement and Muirhead $\sum_{cyc}(x^4y+x^4yz+x)\geq3\sum_{cyc}\sqrt[3]{x^9y^2z}=3\sqrt[3]{x^2y^2z^2}\sum_{cyc}\sqrt[3]{\frac{x^7}{z}}\geq$ $\geq3\sqrt[3]{x^2y^2z^2}\sum_{cyc}\sqrt[3]{\frac{x^7}{x}}=3\sum_{cyc}\sqrt[3]{x^8y^2z^2}\geq3\sum_{cyc}\sqrt[3]{x^6y^3z^3}=3\sum_{cyc}x^2yz$.

crazyfehmy wrote: Prove that \[ x^4y+y^4z+z^4x+xyz(x^3+y^3+z^3) \geq (x+y+z)(3xyz-1) \]for all positive real numbers $x, y, z$. By AM-GM , $\sum_{cyc}x^4y+xyz\sum_{cyc}x^3\ge \sum_{cyc}x^4y+xyz\sum_{cyc}xy^2=\sum_{cyc}(x^4y+xy^2z^3+x)-\sum_{cyc}x\ge 3\sum_{cyc}x^2yz-\sum_{cyc}x=(3xyz-1)\sum_{cyc}x.$

http://artofproblemsolving.com/community/c6h1198039p5876818

Eray wrote: http://artofproblemsolving.com/community/c6h1198039p5876818 Thanks.

crazyfehmy wrote: Prove that \[ x^4y+y^4z+z^4x+xyz(x^3+y^3+z^3) \geq (x+y+z)(3xyz-1) \]for all positive real numbers $x, y, z$. Suffice instead of the objective to prove $\displaystyle {x ^ 4 y + y ^ 4 z + z ^ 4 x + x + y + z + xyz (x ^ 3 + y ^ 3 + z ^ 3) \geq 3xyz (x + y + z)},$ or is$ 6 \root 6 \of {{x ^ 6} {y ^ 6} {z ^ 6}} + xyz \left ({{x ^ 3} + {y ^ 3} + {z ^ 3}} \right) - 3xyz (x + y + z) \geqslant 0 $ or is $6+ {x ^ 3} + {y ^ 3} + {z ^ 3} - 3x-3y-3z \geqslant 0,$ so long as he can prove that$ {\left ({x - 1} \right)^ 2} \left ({x + 2} \right) + {\left ({y - 1} \right) ^ 2} \left ({y + 2} \right) + {\left ({z - 1} \right) ^ 2} \left ({z + 2} \right) \geqslant 0,$ applicable to any triad of positive real numbers $ x, y, z. $ Gender certainly true for$ x = y = z = 1.$

$x^4y+y^4z+z^4x+xyz(x^3+y^3+z^3)+x+y+z \geq 3xyz(x+y+z)$ $x^4y+y \geq 2x^2y$ $y^4z+z \geq 2y^2z$ $z^4x+x \geq 2z^2x$ adding these inequalities $xyz(2(\frac{x}{z}+\frac{y}{x}+\frac{z}{y})+x^3+y^3+z^3) \geq 3xyz(x+y+z)$ $\frac{x}{z}+\frac{y}{x}+\frac{z}{y} \geq 3$ Then, $x^3+1+1+y^3+1+1+z^3+1+1 \geq 3(x+y+z)$ indeed true.

$$\sum (x^4y+x^4yz+x)\geq 3\sqrt[3]{xyz} \sum \sqrt[3]{x^8y} \geq 3\sqrt[3]{xyz} \sum \sqrt[3]{x^5y^2z^2}=3(x+y+z)xyz$$

\[\sum{\frac{x^3}{z}}+\sum{\frac{1}{xy}}+\sum{x^3}\geq 3+3+\frac{(\sum{x})^3}{9}\geq 3\sum{x}\]Multiplying both sides by $xyz$ gives the desired result.