orl wrote: Prove the inequality \[ {x\over y^2-z}+{y\over z^2-x}+{z\over x^2-y} > 1, \] where $2 < x, y, z < 4.$ Proposed by A. Golovanov $\sum_{cyc}\frac{x}{y^2-z}\geq\frac{(\sqrt x+\sqrt y+\sqrt z)^2}{\sum\limits_{cyc}(x^2-x)}$. Hence, we need to prove that $f(x,y,z)>0$, where $f(x,y,z)=\sum_{cyc}(2x^2+2xy-x^4)$ and $\{x,y,z\}\subset(\sqrt2,2)$. But $\frac{\partial^2f}{\partial x^2}=4-12x^2<0$ and similarly $\frac{\partial^2f}{\partial y^2}<0$ and $\frac{\partial^2f}{\partial z^2}<0$. Hence, $\inf f=\min\{f\left(\sqrt2,\sqrt2,\sqrt2\right),f\left(\sqrt2,\sqrt2,2\right),f\left(\sqrt2,2,2\right),f\left(2,2,2\right)\}=$ $=f(2,2,2)=0$.

Let's prove that $z(y^2-z)<3y^2$: Because $z<4$, it is enough to prove that the least root of the equation $z^2-zy^2+3y^2=0$ in $z$ is greater than $4$, what is true, because: $\frac{y^2-\sqrt{y^4-12y^2}}{2}>4 \Leftrightarrow (y^2-8)^2> y^4-12y^2 \Leftrightarrow y < 4$. (we can square the inequality because $y^2>8$, otherwise the determitant is negative and the inequality is clearly true) Now, $\sum \frac{x}{y^2-z}=\sum \frac{xz}{z(y^2-z)} > \sum \frac{xz}{3y^2} \ge 1$ (by AM-GM).

rsa365 your solution is nice and simple! $ z(y^2 -z) < 3y^2 $ is proved more easily. Use AM-GM and 4>y,z ${ 3y^2 +z^2 \geq 4 \sqrt {y^3 z} = \sqrt {4y^3} \sqrt {4z} > \sqrt {y^4} \sqrt {z^2} = y^2 z} $.

orl wrote: Prove the inequality \[ {x\over y^2-z}+{y\over z^2-x}+{z\over x^2-y} > 1, \] where $2 < x, y, z < 4.$ Proposed by A. Golovanov From the condition we have: $(4-y)(y-2) > 0 \Leftrightarrow y^2 < 6y-8$. Therefore, using this fact and Cauchy-Schwarz, we have: $\sum \frac{x}{y^2-z} > \sum \frac{x}{6y-8-z} \ge \frac{(\sum x )^2}{5\sum xy-8\sum x}$ Hence, it remains to prove: $\sum x^2+8\sum x \ge 3\sum xy$ What follows from $\sum xy \leq \sum x^2$ and $4x> x^2$.