Who's up to the challenge? 7

By Holder's Inequality, we have:

$\displaystyle \sum_{i=1}^m { { x }_{ i }^{ 4 } } \cdot \displaystyle \sum_{i=1}^m { { y }_{ i }^{ \frac{4}{3} } } \cdot \displaystyle \sum_{i=1}^m { { y }_{ i }^{ \frac{4}{3} } }\cdot \displaystyle \sum_{i=1}^m { { y }_{ i }^{ \frac{4}{3} } } \geq (\displaystyle\sum _{ i=1 }^{ m }{ { x }_{ i }{ y }_{ i } })^4$

$\Rightarrow (1296 \times 16 \times 16 \times 16)^{\frac{1}{4}} \geq \displaystyle\sum _{ i=1 }^{ m }{ { x }_{ i }{ y }_{ i } }$

$\Rightarrow 48 \geq \displaystyle\sum _{ i=1 }^{ m }{ { x }_{ i }{ y }_{ i } }$

This is the exact statement of Holder's Inequality and a special case of it yields the maximum value of the required summation to be $(1296^{1/4})*(16^{1-\frac{1}{4}})=6*8=48$

By Holder's inequality $\left(\sum_{i=1}^m x_i^4\right)^{\frac 14}\left(\sum_{i=1}^m y_i^{\frac 43}\right)^{\frac 34}\ge \sum_{i=1}^m x_iy_i$