An inequality on a convex function

It all begins with the Cauchy-Schwarz inequality which says that the average of the squares is bigger than the square of the average: $$\left({x+y\over 2}\right)^2\leq {x^2+y^2\over 2}.$$ We see this is true since subtracting the left from the right hand expression gives $(x-y)^2/4\geq 0$.

Next, we extend this result to collections larger than two:
$$\left({x_1+x_2+\cdots +x_n\over n}\right)^2\leq {x_1^2+x_2^2+\cdots+x_n^2\over n}.$$ You can prove this by induction on $n$, or directly by subtracting the left from the right hand expression to get the obviously non-negative $\sum_{i<j} (x_i-x_j)^2/n^2$.

Now, applying this inequality twice, once to the numbers $x_j$ and then to the numbers $x_j^2$, gives the fourth power version: $$\left({x_1+x_2+\cdots+x_n\over n}\right)^4\leq \left({x_1^2+x_2^2+\cdots+x_n^2\over n}\right)^2 \leq {x_1^4+x_2^4+\cdots+x_n^4\over n}. $$

Finally, as mentioned by Carl, putting $n=12$ and choosing $x_j=a$ for $j=1,2,3,4,5,6$, $x_j=b$ for $j=7,8,9,10$, $x_{11}=c$, and $x_{12}=d$ then plugging into the formula above gives your result.

I don't know how well this would go over in a grade 8 classroom, but it doesn't use any advanced mathematics.

Let $\{a_j\}$ be a set of $n$ numbers. Then the 4th power of the average of $\{a_j\}$ is smaller than the average of the 4th powers of $\{a_j\}$: $$(\Sigma_ja_j/n)^4 \le \Sigma_j a_j^4/n.$$ With $n=12$ and appropriate choice of $a_j$ this will give you what you want.

Unfortunately, being an old foggy, I can't recall why it is that I know the above fact, but it should be easy to show with calculus.