Inequality of sum I am trying to solve a inequality consisting of the sums.
QUESTION:
$\text{Suppose that } a_1, \dots, a_n\ \in \mathbb{R}. \text{Show that:}$
$$\left(\frac{1}{n}\sum _{k=1}^na_k^2\right)^{\frac{1}{2}}\le \left(\frac{1}{n}\sum _{k=1}^na_k^4\right)^{\frac{1}{4}}$$
WHAT I HAVE DONE SO FAR:
I substitute $a_k^2=b_k$ and raise the entire inequality to the power of two giving me:
$$\left(\sum _{k=1}^nb_k\right)^2\le n\sum _{k=1}^nb_k^2$$
And since the left side of the inequality consists of the terms $b_k^2$ and the terms $2b_ib_j$, we can subtract n of the square terms from both sides of the inequality:
$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2+nb_2^2+nb_3^2+...+nb_n^2-b_1^2-b_2^2-b_3^2-...-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2-b_1^2+nb_2^2-b_2^2+nb_3^2-b_3^2+...+nb_n^2-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+\left(n-1\right)b_3^2+...+\left(n-1\right)b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)\sum _{k=1}^nb_k^2$$
But after this I seem to get stuck. According the answer-sheet, I am supposed to get here from the previous expression:
$$0\le \sum_{i<j}(b_i-b_j)^2$$
which would mean the original inequality is true, because these are equivalent. Can someone please explain to me why that is. Any help would be very much appreciated!
 A: After some more thinking, I managed to figure it out:
SOLUTION:
I substitute $a_k^2=b_k$ and raise the entire inequality to the power of two giving me:
$$\left(\sum _{k=1}^nb_k\right)^2\le n\sum _{k=1}^nb_k^2$$
And since the left side of the inequality consists of the terms $b_k^2$ and the terms $2b_ib_j$, we can subtract n of the square terms from both sides of the inequality:
$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2+nb_2^2+nb_3^2+...+nb_n^2-b_1^2-b_2^2-b_3^2-...-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le nb_1^2-b_1^2+nb_2^2-b_2^2+nb_3^2-b_3^2+...+nb_n^2-b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+\left(n-1\right)b_3^2+...+\left(n-1\right)b_n^2$$
$$\sum _{i<j}^{ }2b_ib_j\le \left(n-1\right)\sum _{k=1}^nb_k^2$$
We rearrange the inequality:
$$0\le \left(n-1\right)\sum _{k=1}^nb_k^2-\sum _{i<j}^{ }2b_ib_j$$
Which when expanded becomes:
$$0\le \left(n-1\right)\left(b_1^2+b_2^2+...b_n^2\right)-\left(2b_1b_2+2b_1b_3+....+2b_{n-1}b_n\right)$$
$$0\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+...\left(n-1\right)b_n-\left(2b_1b_2+2b_1b_3+....+2b_{n-1}b_n\right)$$
$$0\le \left(n-1\right)b_1^2+\left(n-1\right)b_2^2+...+\left(n-1\right)b_n-2b_1b_2-2b_1b_3-....-2b_{n-1}b_n$$
To make a simple conclusion from this, I might just decide to see what happens when $n=3$.  Then the inequality would become:
$$0\le 2b_1^2+2b_2^2+2b_3^2-2b_1b_2-2b_1b_3-2b_2b_3$$
Which is the same as:
$$0\le \left(b_1^2-2b_1b_2+b_2^2\right)+\left(b_1^2-2b_1b_3+b_3^2\right)+\left(b_2^2-2b_2b_3+b_3^2\right)$$
$$0\le \left(b_1-b_2\right)^2+\left(b_1-b_3\right)^2+\left(b_2-b_3\right)^2$$
And from here I would make the conclusion that:
$$0\le\sum_{i<j}^{ }\left(b_i-b_j\right)^2$$
And since this inequality is equivalent with the original inequality, we would then know that the original inequality is true.
