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:
$$\left(y_1^2+{\cdots}+y_p^2\right)^2\leq{}\left(y_1^2+{\cdots}+y_p^2\right)\cdot{}\left(y_1^2+{\cdots}+y_p^2\right)$$

Further hints: 
(1) Can you find a proof for $i = 1$ and $p = 2$, i.e., can you show that 
$$
\left(\frac{x+y}{2}\right)^2 < \frac{x^2 + y^2}{2},
$$
provided that $x \ne y$, and both $x$ and $y$ are positive? How does $x \ne y$ figure in your proof? 
(2) In the restated goal, you have a bunch of numbers raised to the $i$th power, and the $2i$th power. How about letting $x_j = (y_j)^i$, so that $x_j^2 = (y_j)^{2i}$. Does that help simplify things a bit? 
To respond to the comment ("I can't figure out #1"), I'm going to write this one out. Of course, if you can't write this one out, there's no hope of doing the more general proof; it's one of the reasons I often try to prove an easy case of a theorem before moving on to the general case. Here goes:
We want to show that if $x$ and $y$ differ, then 
\begin{align}
\left(\frac{x+y}{2}\right)^2 &< \frac{x^2 + y^2}{2},
\end{align}
I'll do so by sequentially altering the inequality to equivalent inequalities:
\begin{align}
\left(\frac{x+y}{2}\right)^2 &< \frac{x^2 + y^2}{2}\\
\frac{x^2+2xy+y^2}{4} &< \frac{x^2 + y^2}{2} & \text{by algebra}\\
x^2+2xy+y^2 &< 2x^2 + 2y^2 & \text{multiply through by 4}\\
2xy&< x^2 + y^2 & \text{subtract $x^2 + y^2$ from both sides}\\
0&< x^2 -2xy + y^2 & \text{subtract $2xy$ from both sides}\\
0&< (x-y)^2& \text{factor right hand side}\\
\end{align}
Now observe that since $x \ne y$, the right hand side is nonzero. 
If you read @robjohn's answer, you'll see that it's just those steps, written out in reverse order and generalized to a sum of $p$ items, I believe. 
