# Minkovski's Inequality

• Jun 6th 2006, 01:54 PM
Judi
Minkovski's Inequality
We were given this inequality to prove:

sqrt(summation of (ak + bk)^2 from K=1 to n) <=

sqrt(summation of (ak)^2 from K=1 to n) + sqrt(summation of (bk)^2 from K=1 to n)

We were told to square both sides and then use the Cauchy-Schwarz inequality to finish.
• Jun 6th 2006, 02:44 PM
ThePerfectHacker
Quote:

Originally Posted by Judi
We were given this inequality to prove:

sqrt(summation of (ak + bk)^2 from K=1 to n) <=

sqrt(summation of (ak)^2 from K=1 to n) + sqrt(summation of (bk)^2 from K=1 to n)

We were told to square both sides and then use the Cauchy-Schwarz inequality to finish.

You need to show,
$0\leq \sqrt{ \sum_{k=1}^n (ak+bk)^2 }\leq \sqrt{ \sum_{k=1}^n(ak)^2} +\sqrt{ \sum_{k=1}^n(bk)^2 }$
Upon squaring we have, if and only if
$\sum_{k=1}^n (ak+bk)^2\leq \sum_{k=1}^n(ak)^2+\sum_{k=1}^n(bk)^2+2\sqrt{\sum_ {k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2$
If and only If
$\sum_{k=1}^n (ak)^2+2(ak)(bk)+(bk)^2\leq$ $\sum_{k=1}^n(ak)^2+\sum_{k=1}^n(bk)^2+2\sqrt{\sum_ {k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2$
if and only if
$2\sum_{k=1} (ak)(bk)\leq 2\sqrt{\sum_{k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2$
if and only if
$\sum_{k=1}^n (ak)(bk) \leq \sqrt{\sum_{k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2$
If and only if *)
$\left( \sum_{k=1}^n (ak)(bk) \right)^2 \leq \sum_{k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2$
This is true by Cauchy-Swarchz Inequality.
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Explanation of last step.
We will prove a famous case,
$\left( \sum_{k=1}^n a_k b_k\right)^2\leq \left(\sum_{k=1}^n a_k^2 \right) \cdot \left( \sum_{k=1}^n b_k^2 \right)$
Because this is true only when, (take square roots)
$\left| \sum_{k=1}^n a_k b_k \right| \leq \sqrt{\sum_{k=1}^n a_k^2}\cdot \sqrt{\sum_{k=1}^n b_k^2 \right)$

Consider vectors in $\mathbb{R}^n$
$\bold{v}=(a_1,a_2,...,a_n)$
$\bold{u}=(b_1,b_2,...,b_n)$
Then,
$|\bold{v} \cdot \bold{u}|\leq ||\bold{v}||\cdot ||\bold{u}||$-by Cauchy-Swartchz
But,
$|\bold{v}\cdot \bold{u}|=\left| \sum_{k=1}^n a_k b_k \right|$
and,
$||\bold{v}||\cdot ||\bold{u}||=\sqrt{\sum_{k=1}^n a_k^2}\cdot \sqrt{\sum_{k=1}^n b_k^2 \right)$
$\mathbb{Q}.\mathbb{E}.\mathbb{D}$

*)If LHS is negative there is nothing to prove because RHS must be positive! We assume it is positive also.

Note: The consistent use of 'if and only if' otherwise the proof falls.