# Thread: Help with 2 Other Questions ??

1. ## Help with 2 Other Questions ??

i If someone sends an answer
please write that in mathematical form

i have problems with my english

2. For any $\displaystyle n>0$

Consider the vector space $\displaystyle \mathbb{R}^n$ over $\displaystyle \mathbb{R}$.

Now, let
$\displaystyle \bold{u}=\left<1,\frac{1}{2},\frac{1}{3},...,\frac {1}{n}\right>$

$\displaystyle \bold{v}=\left<1,\frac{1}{2^3},\frac{1}{3^3},...,\ frac{1}{n^3}\right>$

Then by Cauchy-Schwartz Inequality we have,
$\displaystyle \bold{ab}\leq ||\bold{a}||||\bold{b}||$

Now, (the dot product)
$\displaystyle \bold{ab}=1+\frac{1}{2^4}+...\frac{1}{n^4}$

And the norms,
$\displaystyle ||\bold{a}||=\sqrt{1+\frac{1}{2^2}+...+\frac{1}{n^ 2}}$
$\displaystyle ||\bold{b}||=\sqrt{1+\frac{1}{2^6}+...+\frac{1}{n^ 6}}$
Thus,

$\displaystyle 1+\frac{1}{2^4}+...+\frac{1}{n^4}\leq \sqrt{1+\frac{1}{2^2}+...+\frac{1}{n^6}}\cdot \sqrt{1+\frac{1}{2^6}+...+\frac{1}{n^6}}$

Square both sides (non-negative terms),
$\displaystyle \left(1+\frac{1}{2^4}+...+\frac{1}{n^4}\right)^2\l eq$$\displaystyle \left(1+\frac{1}{2^2}+...+\frac{1}{n^2}\right)\cdo t$$\displaystyle \left(1+\frac{1}{2^6}+...+\frac{1}{n^6}\right)$

3. Let,
$\displaystyle x\geq 1$
Then,
$\displaystyle \frac{1}{x^n}-\frac{1}{x^{n+1}}=\frac{x^{n+1}-x^n}{x^{2n+1}}\leq x^{n+1}-x^n$
Thus,
$\displaystyle x^{n+1}-x^n\geq \frac{1}{x^n}-\frac{1}{x^{n+1}}$
Thus,
$\displaystyle x^{n+1}+\frac{1}{x^{n+1}}\geq x^n+\frac{1}{x^n}$

Now, if $\displaystyle 0<x<1$
Then,
$\displaystyle \frac{1}{x}>1$
Thus, (by the previous discussion),
$\displaystyle (1/x)^{n+1}+\frac{1}{(1/x)^{n+1}}\geq (1/x)^n+\frac{1}{(1/x)^n}$
Thus,
$\displaystyle x^{n+1}+\frac{1}{x^{n+1}}\geq x^n+\frac{1}{x^n}$