49th Austrian Mathematical Olympiad, National Competition (Final Round, part 1)

By replacing $\alpha$ by $xy+yz+zx$ and clearing fractions we get the equivalent inequality $$(x^2+xy+xz+yz)(y^2+yx+yz+xz)(z^2+zx+zy+xy)\ge Cx{y}^{2}z(x^2+z^2+2xz).$$ This inequality is homogeneous of degree 6, thus no further constraint has to be considered. As each of the three factors on the left-hand-side can be factorized we get $$(x+y)(x+z)(y+x)(y+z)(z+x)(z+y)\ge Cx{y}^{2}z(x+z{)}^2.$$ Upon cancellation of $(x+z{)}^2$ we arrive at the equivalent inequality $$(x+y{)}^2(z+y{)}^2\ge Cx{y}^{2}z.$$ Estimating each of the two factors on the left with the arithmetic-geometric inequality we obtain the optimal value of $C=16$ which is attained by $x=y=z=\sqrt{\alpha }/3$.