Austria 2010

We have $$\begin{array}{rl}\frac{(3x-1{)}^2}{x}+\frac{(3y-1{)}^2}{y}& =\frac{9x^2-6x+1}{x}+\frac{9y^2-6y+1}{y}\\ & =9x-6+\frac{1}{x}+9y-6+\frac{1}{y}\\ & =-3+\frac{1}{x}+\frac{1}{y}.\end{array}$$ It remains to show that $$\frac{1}{x}+\frac{1}{y}\ge 4.$$ This is a consequence of the inequality between the arithmetic and the harmonic mean and the condition $x+y=1$: $$(x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\ge 4.$$ Equality holds exactly for $x=y$ and therefore for $x=y=1/2$.