Balkan Mathematical Olympiads

The given condition can be rearranged to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$. Using this, we obtain: $$\begin{array}{rl}{x}^{2}y+y^{2}z+z^{2}x-2(x+y+z)-3& =x^{2}y-2x+\frac{1}{y}+y^{2}z-2y+\frac{1}{z}+z^{2}x-2z+\frac{1}{x}\\ & =y{\left(x-\frac{1}{y}\right)}^2+z{\left(y-\frac{1}{z}\right)}^2+x{\left(z-\frac{1}{x}\right)}^2\ge 0\end{array}$$ Equality holds if and only if we have $xy+yz+zx=1$, or, in other words, $x=y=z=1$.

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Alternative solution.

It follows from $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$ and Cauchy-Schwarz inequality that $$\begin{array}{rl}3(x^{2}y+y^{2}z+z^{2}x)& =\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x^{2}y+y^{2}z+z^{2}x)\\ & =\left({\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2\right)\left((x\sqrt{y}{)}^2+(y\sqrt{z}{)}^2+(z\sqrt{x}{)}^2\right)\ge \\ & =(x+y+z{)}^2\end{array}$$ Therefore, $x^{2}y+y^{2}z+z^{2}x\ge \frac{(x+y+z{)}^2}{3}$ and if $x+y+z=t$ it suffices to show that $\frac{t^2}{3}\ge 2t-3$. The latter is equivalent to $(t-3{)}^2\ge 0$. Equality holds when $$x\sqrt{y}\sqrt{y}=y\sqrt{z}\sqrt{z}=z\sqrt{x}\sqrt{x}$$ i.e. $xy=yz=zx$, and $t=x=y=z=3$. Hence, $x=y=z=1$.