crazyfehmy wrote: Prove that \[ (x^4+y)(y^4+z)(z^4+x) \geq (x+y^2)(y+z^2)(z+x^2) \]for all positive real numbers $x, y, z$ satisfying $xyz \geq 1$. We need to prove that $x^4y^4z^4+\sum_{cyc}(x^5y^4+x^5z)\geq x^2y^2z^2+\sum_{cyc}(x^3z^2+x^3y)$, which is true because $x^4y^4z^4\geq x^2y^2z^2$ and by AM-GM $\sum_{cyc}(x^5y^4+2z^5x^4)\geq3\sum_{cyc}x^3z^2$ and $\sum_{cyc}(2x^5z+y^5x)\geq3\sum_{cyc}x^3y$. Done!

There is also a solution without expanding and only using Cauchy-Schwarz.

crazyfehmy wrote: There is also a solution without expanding and only using Cauchy-Schwarz. Can you write this solution ?

Nice Problem By Cauchy-Schwarz we have that $(x^4+y)(\frac{z^2}{y}+1)\geq (x^2+z)^2$ $\to$ $(x^4+y)(z^2+y)\geq y(x^2+z)^2$ Similarly \[(y^4+z)(x^2+z)\geq z(y^2+x)^2\]\[(x^4+y)(z^2+y)\geq y(x^2+z)^2\]\[(z^4+x)(y^2+x)\geq x(z^2+y)^2.\]$\to$\[ (x^4+y)(y^4+z)(z^4+x) \geq xyz(x+y^2)(y+z^2)(z+x^2)\geq (x+y^2)(y+z^2)(z+x^2) \]Since $xyz\geq 1$

IstekOlympiadTeam wrote: Nice Problem By Cauchy-Schwarz we have that (x^4+y)(\frac{z^2}{y}+1)\geq (x^2+z)^2$ $\to$ $(x^4+y)(z^2+y)\geq y(x^2+z)^2$ Similarly (y^4+z)(x^2+z)\geq z(y^2+x)^2(x^4+y)(z^2+y)\geq y(x^2+z)^2(z^4+x)(y^2+x)\geq x(z^2+y)^2. $\to$ (x^4+y)(y^4+z)(z^4+x) \geq xyz(x+y^2)(y+z^2)(z+x^2)\geq (x+y^2)(y+z^2)(z+x^2) Since $xyz\geq 1$ effective solution thank a lot ISEKOLMPIA

Let \(L\) and \(R\) denote the left- and right-hand expressions. Expanding, \[L=x^4y^4z^4+\sum_\mathrm{cyc} x^5y^4+\sum_\mathrm{cyc} x^5y+xyz\quad\text{and}\quad R=x^2y^2z^2+\sum_\mathrm{cyc} x^3z^2+\sum_\mathrm{cyc} x^3y+xyz.\]Subtracting, \[L-R=\Big(x^4y^4z^4-x^2y^2z^2\Big)+\left(\sum_\mathrm{cyc} x^5y^4-\sum_\mathrm{cyc} x^3z^2\right)+\left(\sum_\mathrm{cyc} x^5z-\sum_\mathrm{cyc} x^3y\right).\]First, \(x^4y^4z^4\ge x^2y^2z^2\) is obvious. Second, note that \[\frac{2z^5x^4+x^5y^4}3\ge x^{13/3}y^{4/3}z^{10/3}\ge x^3z^2,\]and summing cyclically, we have \(\sum_\mathrm{cyc} x^5y^4\ge\sum_\mathrm{cyc} x^3z^2\). Third, note that \[\frac{y^5x+2x^5z}3\ge x^{11/3}y^{5/3}z^{2/3}\ge x^3y,\]and summing cyclically, we have \(\sum_\mathrm{cyc} x^5z\ge\sum_\mathrm{cyc} x^3t\). Hence all the bracketed terms are nonzero, so \(L-R\ge0\), as desired.