Help with 2 Other Questions ??

**sbsite** · November 2nd 2006, 12:56 AM

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

i have problems with my english

**ThePerfectHacker** · November 2nd 2006, 05:35 AM

For any $n>0$

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

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

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

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

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

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

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

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

**ThePerfectHacker** · November 2nd 2006, 05:55 AM

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

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

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