But it's not symmetrical!

[This is not a complete solution. You should be able to fill in the details]

Consider 3 vectors with lengths $x, y,$ and $z$ that are $120^ \circ$ apart. (If the lengths are negative, it means that the vector is in the opposite direction.)

1. What does the conditions $x^2 + xy + y^2 = 3 , y^2 + yz + z^2 = 16$ tell us about the triangle $XYZ$ ?
2. How is the expression $xy+yz+zx$ related to triangle $XYZ$ ?
3. Hence, how can we calculate the maximum value of $M = xy + yz + zx$ ?
   Hint: $\frac{1}{2} M \times \frac{ \sqrt{3} } { 2} \leq \frac{1}{2} \times \sqrt{ 3 \times 16}$ .
   
4. When is the maximium achieved? (In particular, we do not need to calcualte the exact values of $x, y,$ and $z$ .)
5. Generalize!

$(x^2+y^2+xy)(y^2+yz+z^2)$

$=[{ (y+\frac { x }{ 2 } })^{ 2 }+\frac { 3 }{ 4 } { x }^{ 2 }][{ (y+\frac { z }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } { z }^{ 2 }]$

$\ge (\frac { \sqrt { 3 } }{ 2 } )^{ 2 }[z(y+\frac { x }{ 2 } )+x(y+\frac { z }{ 2 } )]^{ 2 }$

$=\frac { 3 }{ 4 } (xy+yz+zx)^{ 2 }$

$=> (xy+yz+zx)\le \left| xy+yz+zx \right| \le \frac { \sqrt { 48*4 } }{ \sqrt { 3 } } =8$

Please tell me if this solution has any problems and/or missing something. Thanks!