Cauchy-Schwarz problem (Maybe) I believe that this may require Cauchy-Schwarz
$$|\vec{u}\cdot\vec{v}|\leq |\vec{u}|\cdot|\vec{v}|$$
to solve.
Let $y_1,y_2,\ldots ,y_p$ be $p$ positive numbers and let $i$ be a positive integer. Let $Z_i$ be the number defined by the formula 
$$Z_i:=\left(\frac{y_1^i+{\cdots}+y_p^i}{p}\right)^{\frac{1}{i}}.$$
Show that if $y_1,y_2,\ldots ,y_p$ are not all equal, then $Z_i$ is strictly less than $Z_{2i}$.
Edit
$$\left(\frac{y_1^i+{\cdots}+y_p^i}{p}\right)^2<\frac{y_1^{2i}+{\cdots}+y_p^{2i}}{p}$$
(raise each side to the $2i$ power.)
Setting $u_k=v_k=y_k$ results in the following statement:
$$\left(y_1^2+{\cdots}+y_p^2\right)^2\leq{}\left(y_1^2+{\cdots}+y_p^2\right)\cdot{}\left(y_1^2+{\cdots}+y_p^2\right)$$

Here is a standard proof of Cauchy-Schwarz that shows why equality only holds when all the terms are the same.

Let $x_k=y_k^i$ and define
$$
\bar x=\frac1p\sum_{k=1}^px_k\tag{1}
$$
Then
$$
\begin{align}
\frac1p\sum_{k=1}^p(x_k-\bar x)^2
&=\frac1p\sum_{k=1}^px_k^2-\frac2p\sum_{k=1}^px_k\bar x+\frac1pp{\bar x}^2\\
&=\frac1p\sum_{k=1}^px_k^2-2\bar x\bar x+{\bar x}^2\\
&=\frac1p\sum_{k=1}^px_k^2-{\bar x}^2\\
&=\frac1p\sum_{k=1}^px_k^2-\left(\frac1p\sum_{k=1}^px_k\right)^2\tag{2}
\end{align}
$$
Note that the left side of $(2)$ is positive, giving strict inequality, unless $x_k=\bar x$ for all $k$.
