Cauchy-Schwarz problem (Maybe)

Question

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:

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

Answer 1

$$ \vec{u}=\left(\begin{array}{c} \frac{x_1^p}{n}\\ \vdots\\ \frac{x_1^p}{n} \end{array} \right), \vec{v}=\left(\begin{array}{c} 1_1\\ \vdots\\ 1_n \end{array} \right) $$ \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*}

Answer 2

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

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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 $k$.

Answer 3

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.