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:
$$\left(y_1^2+{\cdots}+y_p^2\right)^2\leq{}\left(y_1^2+{\cdots}+y_p^2\right)\cdot{}\left(y_1^2+{\cdots}+y_p^2\right)$$
 A: 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 $k$.
A: 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. 
A: $$
\vec{u}=\left(\begin{array}{c}
\frac{x_1^p}{n}\\
\vdots\\
\frac{x_1^p}{n}
\end{array}
\right),
\vec{v}=\left(\begin{array}{c}
1_1\\
\vdots\\
1_n
\end{array}
\right)
$$
\begin{align*}
\Vert\vec{u}\cdot\vec{v}\Vert &\leq\Vert\vec{u}\Vert\cdot\Vert\vec{v}\Vert\\
\frac{x_1^p}{n}+{\cdots}+\frac{x_n^p}{n} &\leq\sqrt{\frac{x_1^{2p}}{n^2}+{\cdots}+\frac{x_n^{2p}}{n^2}}\cdot\sqrt{n}\\
\frac{x_1^p+{\cdots}+x_n^p}{n} &\leq\sqrt{\frac{x_1^{2p}+{\cdots}+x_n^{2p}}{n}}\\
\left(\frac{x_1^p+{\cdots}+x_n^p}{n}\right)^{\frac{1}{p}} &\leq\left(\frac{x_1^{2p}+{\cdots}+x_n^{2p}}{n}\right)^{\frac{1}{2p}}
\end{align*}
