My solution: Observe $$\text{LHS} \ge \sum_{cyc} \frac{2xy+z}{x^2+2} = \sum_{cyc} \frac{2+z^2}{(x^2+2)z} \geq 3 \sqrt[3] {\frac{1}{xyz}} =3$$as desired.

Equality happens at $x=y=z=1$

realquarterb wrote: My solution: Observe $$\text{LHS} \ge \sum_{cyc} \frac{2xy+z}{x^2+2} = \sum_{cyc} \frac{2+z^2}{(x^2+2)z} \geq 3 \sqrt[3] {\frac{1}{xyz}} =3$$as desired. realquarterb wrote: Equality happens at $x=y=z=1$ Nice solution!

IMO2019 wrote: Nice solution! Indeed, but I think inequality is kind of weak. (Usually contest inequalities are not too strong) Look at this fb page tho:[url] https://www.facebook.com/groups/TAoMathematics/ [/url]for some very insane inequalities.

Inequality might be weak but the bound is strong!

Amir Hossein wrote: Let $x,y,z$ be positive reals such that $xyz=1$. Show that $$\frac{x^2+y^2+z}{x^2+2} + \frac{y^2+z^2+x}{y^2+2} + \frac{z^2+x^2+y}{z^2+2} \geq 3.$$When does equality happen? Let $x,y,z$ be positive reals such that $xyz=1$. For $\lambda\geq 0,$ show that $$\frac{x^2+y^2+z}{x^2+\lambda} + \frac{y^2+z^2+x}{y^2+\lambda} + \frac{z^2+x^2+y}{z^2+\lambda} \geq \frac{9}{1+\lambda}.$$here