European Mathematical Cup

a) Use AM-GM and $xyz=8$ to get $$x^2+y^2+z^2+xy+xy+xz+xz+yz+yz\ge 9\sqrt{x^6{y}^6{z}^6}=36.$$

We have equality for $x=y=z=2$.

b) Using $xyz=8$, we can transform the given expression: $$x^2+y^2+z^2+mxy+nxz+pyz=x^2+\frac{8p}{x}+y^2+\frac{8n}{y}+z^2+\frac{8m}{z}.$$ Since all numbers are positive reals, we can apply AM-GM inequality to get: $$x^2+\frac{8p}{x}=x^2+\frac{4p}{x}+\frac{4p}{x}\ge 6\sqrt{p^2}.$$ When we apply the same procedure for $x,y,z$ and sum the inequalities, we get: $$x^2+y^2+z^2+mxy+nxz+pyz=x^2+\frac{8p}{x}+y^2+\frac{8n}{y}+z^2+\frac{8m}{z}\ge 6\sqrt[3]{2}(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2}).$$ In order to get equality, we must have equality in all above inequalities and that happens for $$x=\sqrt[3]{4p},y=\sqrt[3]{4n},z=\sqrt[3]{4m}.$$

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Alternative solution.

We only present solution for b) part here, marking scheme for a) part is the same as in first solution. We use weighted AM-GM: $$\begin{array}{rlrlrlrl}{x}^2+y^2+z^2+mxy+nxz+pyz& =\sqrt[3]{p^2}\frac{x^2}{\sqrt[3]{p^2}}+\sqrt[3]{n^2}\frac{y^2}{\sqrt[3]{n^2}}+\sqrt[3]{m^2}\frac{z^2}{\sqrt[3]{m^2}}+2\sqrt[3]{m^2}\frac{mxy}{2\sqrt[3]{m^2}}+2\sqrt[3]{n^2}\frac{nxz}{2\sqrt[3]{n^2}}+2\sqrt[3]{p^2}\frac{pyz}{2\sqrt[3]{p^2}}& \ge 3(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})\cdot 3(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})\sqrt[3]{\frac{x^2}{\sqrt[3]{p^2}}}\sqrt[3]{\frac{y^2}{\sqrt[3]{n^2}}}\sqrt[3]{\frac{z^2}{\sqrt[3]{m^2}}}& =3(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})\sqrt[3]{\frac{(3\sqrt{mxy}{)}^2}{2}}\sqrt[3]{\frac{(3\sqrt{nxz}{)}^2}{2}}\sqrt[3]{\frac{(3\sqrt{pyz}{)}^2}{2}}& =3(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})(\sqrt[3]{\frac{(xyz{)}^2}{2}}{)}^2& =3(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})\sqrt[3]{\frac{(xyz{)}^2}{2}}& =3(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})\sqrt[3]{4^2}& =6\sqrt[3]{2}(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})\end{array}$$ We have shown that the minimum value the expression can take is $6\sqrt[3]{2}(\sqrt[3]{m^2}+\sqrt[3]{n^2}+\sqrt[3]{p^2})$. Equality can only be achieved when $x=\sqrt[3]{4p},y=\sqrt[3]{4n},z=\sqrt[3]{4m}$.