Problem 2. Let $m, n, p$ be fixed positive real numbers which satisfy $m n p=8$. Depending on these constants, find the minimum of

$$
x^{2}+y^{2}+z^{2}+m x y+n x z+p y z
$$

where $x, y, z$ are arbitrary positive real numbers satisfying $x y z=8$. When is the equality attained? Solve the problem for:
a) $m=n=p=2$,

b) arbitrary (but fixed) positive real $m, n, p$.

(Stijn Cambie)
First Solution. a) Use AM-GM and $x y z=8$ to get

$$
x^{2}+y^{2}+z^{2}+x y+x y+y z+y z+x z+x z \geqslant 9 \sqrt[9]{x^{6} y^{6} z^{6}}=36
$$

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

b) Using $x y z=8$, we can transform the given expression:

$$
x^{2}+y^{2}+z^{2}+m x y+n x z+p y z=x^{2}+\frac{8 p}{x}+y^{2}+\frac{8 n}{y}+z^{2}+\frac{8 m}{z}
$$

Since all numbers are positive reals, we can apply AM-GM inequality to get:

$$
x^{2}+\frac{8 p}{x}=x^{2}+\frac{4 p}{x}+\frac{4 p}{x} \geqslant 6 \sqrt[3]{2 p^{2}}
$$

When we apply the same procedure for $x, y, z$ and sum the inequalities, we get:

$$
x^{2}+y^{2}+z^{2}+m x y+n x z+p y z=x^{2}+\frac{8 p}{x}+y^{2}+\frac{8 n}{y}+z^{2}+\frac{8 m}{z} \geqslant 6 \sqrt[3]{2}\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right)
$$

In order to get equality, we must have equality in all above inequalities and that happens for

$$
\begin{aligned}
& x=\sqrt[3]{4 p} \\
& y=\sqrt[3]{4 n} \\
& z=\sqrt[3]{4 m}
\end{aligned}
$$

Desired minimum is therefore

$$
6 \sqrt[3]{2}\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right)
$$

Second 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:

$$
\begin{aligned}
& x^{2}+y^{2}+z^{2}+m x y+n x z+p y z= \\
& \sqrt[3]{p^{2}} \frac{x^{2}}{\sqrt[3]{p^{2}}}+\sqrt[3]{n^{2}} \frac{y^{2}}{\sqrt[3]{n^{2}}}+\sqrt[3]{m^{2}} \frac{z^{2}}{\sqrt[3]{m^{2}}}+2 \sqrt[3]{m^{2}} \frac{m x y}{2 \sqrt[3]{m^{2}}}+2 \sqrt[3]{n^{2}} \frac{n x z}{2 \sqrt[3]{n^{2}}}+2 \sqrt[3]{p^{2}} \frac{p y z}{2 \sqrt[3]{p^{2}}} \geq \\
& 3\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right) \cdot \sqrt[3\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right)]{\left(\frac{x^{2}}{\sqrt[3]{p^{2}}}\right)^{\sqrt[3]{p^{2}}}\left(\frac{y^{2}}{\sqrt[3]{n^{2}}}\right)^{\sqrt[3]{n^{2}}}\left(\frac{z^{2}}{\sqrt[3]{m^{2}}}\right)^{\sqrt[3]{m^{2}}}} \\
& 3\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right) \sqrt{\left(\frac{\sqrt[3]{m} x y}{2}\right)^{2 \sqrt[3]{m^{2}}}\left(\frac{\sqrt[3]{n} x z}{2}\right)^{2 \sqrt[3]{n^{2}}}\left(\frac{\sqrt[3]{p} y z}{2}\right)^{2 \sqrt[3]{p^{2}}}} \\
& =3\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right) \cdot \sqrt[3\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right)]{\left(\frac{x y z}{2}\right)^{2\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right)}} \\
& =3\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right) \cdot \sqrt[3]{\left(\frac{x y z}{2}\right)^{2}}=3\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right) \cdot \sqrt[3]{4^{2}}=6 \sqrt[3]{2}\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right)
\end{aligned}
$$

We have shown that the minimum value the expression can take is $6 \sqrt[3]{2}\left(\sqrt[3]{m^{2}}+\sqrt[3]{n^{2}}+\sqrt[3]{p^{2}}\right)$. Equality can only be achieved when $x=\sqrt[3]{4 p}, y=\sqrt[3]{4 n}, z=\sqrt[3]{4 m}$.