parmenides51 wrote: a) Prove that, for any real $x>0$, it is true that $x^3-3x\ge -2$ . b) Prove that, for any real $x,y,z>0$, it is true that $$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}+2\left(\frac{y}{xz}+\frac{z}{xy}+\frac{x}{yz} \right)\ge 9$$. When we have equality ? https://artofproblemsolving.com/community/c6h1332238p7188341 $$\frac{z^2x}{y} + \frac{x}{yz} + \frac{y}{zx} \geq 3\sqrt[3]{\frac{x}{y}}$$

a) The given inequality is equivalent to: $x^3 + 2 \geq 3x$ So by AM-GM: $x^3+1+1\geq 3\sqrt[3]{x^3}=3x$ as desired The equality occurs if and only if x=1. b)By AM-GM: $\frac {x^2 y} {z} + \frac{y^2 z}{x} +\frac {z^2 x}{y} + \frac {y}{xz} + \frac {z}{xy} + \frac {x}{yz} + \frac {y}{xz} + \frac {z}{xy} + \frac {x}{yz} \geq 9 \sqrt[9] {\frac{(xyz)^5}{(xyz)^5}}=9$ as desired. Equality occurs if and only if x=y=z

a) x^3-3x+2 factors as (x-1)^2(x+2) which is nonnegative for x>0. b) AM-GM over nine terms, equality at x=y=z=1.