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Solve: $|u+v| le |u| + |v|$ with $|x| = left( sqrt + sqrt right)^2$

Vector norms on $Bbb R^n$Prove that $fracx_11+x_2+x_3+ldots+x_n+fracx_21+x_1+x_3+ldots+x_n+ldots+fracx_n1+x_1+x_2+ldots+x_n-1gefracn2n-1$.Prove/Disprove that this is a norm on $mathbbR^3$ spaceWe have the scalar product $left langle cdot, cdot right rangle$. How is its belonging norm $left | cdot right |$ defined?Is $|(I_m + U^top D U)^-1 U^top D|_F$ always small?A Minkowski like inequality for symmetric sumsUse Triangle Inequality to Solve Inequality $left|x+frac12right| > fracsqrt52$Show that the “calculation rule” is wrong for the given scalar productShow that this is not true: $leftlanglevecu,vecvrightrangle^2=vecu^2cdotvecv^2$Need a hint on how to solve this inequality

1

1

$begingroup$

I was given the following task:

Check if $xrightarrow left(sqrt + sqrtright)^2$ is a norm on $mathbbR^2$.

I've already shown that
$$|x| ge 0qquad |x| = 0 Leftrightarrow x = 0$$
$$|alpha x| = |alpha| cdot |x|$$

The last thing that I need to show is the triangle inequality using $sqrtale sqrt + sqrtb$

$$|u+v| = left(sqrt + sqrtright)^2$$
$$le left(sqrt + sqrt+ sqrt + sqrt right) ^2$$
$$= left(sqrt + sqrt+ sqrt + sqrtright)^2$$
$$dots$$

I have no idea how to continue at this point. I've also tried a different approach:
$$|u+v| = left(sqrt + sqrtright)^2$$
$$ge sqrt+^2 = |u_1|+|u_2|+|v_1|+|v_2|$$

but this didn't get me any further as well.
I am thankful for any help

inequality norm examples-counterexamples

share|cite|improve this question

edited Mar 28 at 16:28

TenaliRaman

3,3131224

asked Mar 28 at 14:32

Finn EggersFinn Eggers

413213

$endgroup$

add a comment | 

1

1

$begingroup$

I was given the following task:

Check if $xrightarrow left(sqrt + sqrtright)^2$ is a norm on $mathbbR^2$.

I've already shown that
$$|x| ge 0qquad |x| = 0 Leftrightarrow x = 0$$
$$|alpha x| = |alpha| cdot |x|$$

The last thing that I need to show is the triangle inequality using $sqrtale sqrt + sqrtb$

$$|u+v| = left(sqrt + sqrtright)^2$$
$$le left(sqrt + sqrt+ sqrt + sqrt right) ^2$$
$$= left(sqrt + sqrt+ sqrt + sqrtright)^2$$
$$dots$$

I have no idea how to continue at this point. I've also tried a different approach:
$$|u+v| = left(sqrt + sqrtright)^2$$
$$ge sqrt+^2 = |u_1|+|u_2|+|v_1|+|v_2|$$

but this didn't get me any further as well.
I am thankful for any help

inequality norm examples-counterexamples

share|cite|improve this question

edited Mar 28 at 16:28

TenaliRaman

3,3131224

asked Mar 28 at 14:32

Finn EggersFinn Eggers

413213

$endgroup$

add a comment | 

$begingroup$

I was given the following task:

Check if $xrightarrow left(sqrt + sqrtright)^2$ is a norm on $mathbbR^2$.

I've already shown that
$$|x| ge 0qquad |x| = 0 Leftrightarrow x = 0$$
$$|alpha x| = |alpha| cdot |x|$$

The last thing that I need to show is the triangle inequality using $sqrtale sqrt + sqrtb$

$$|u+v| = left(sqrt + sqrtright)^2$$
$$le left(sqrt + sqrt+ sqrt + sqrt right) ^2$$
$$= left(sqrt + sqrt+ sqrt + sqrtright)^2$$
$$dots$$

I have no idea how to continue at this point. I've also tried a different approach:
$$|u+v| = left(sqrt + sqrtright)^2$$
$$ge sqrt+^2 = |u_1|+|u_2|+|v_1|+|v_2|$$

but this didn't get me any further as well.
I am thankful for any help

inequality norm examples-counterexamples

share|cite|improve this question

edited Mar 28 at 16:28

TenaliRaman

3,3131224

asked Mar 28 at 14:32

Finn EggersFinn Eggers

413213

$endgroup$

share|cite|improve this question

edited Mar 28 at 16:28

TenaliRaman

3,3131224

asked Mar 28 at 14:32

Finn EggersFinn Eggers

413213

share|cite|improve this question

edited Mar 28 at 16:28

TenaliRaman

3,3131224

asked Mar 28 at 14:32

Finn EggersFinn Eggers

413213

edited Mar 28 at 16:28

TenaliRaman

3,3131224

edited Mar 28 at 16:28

TenaliRaman

3,3131224

asked Mar 28 at 14:32

Finn EggersFinn Eggers

413213

asked Mar 28 at 14:32

Finn EggersFinn Eggers

413213

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $u(a^2,b^2)$ and $v(c^2,d^2),$ where $a$, $b$, $c$ and $d$ are positives.

Thus, we need to prove that
$$(a+b)^2+(c+d)^2geqleft(sqrta^2+c^2+sqrtb^2+d^2right)^2$$ or
$$ab+cdgeqsqrt(a^2+c^2)(b^2+d^2),$$ which can be wrong by C-S.

share|cite|improve this answer

answered Mar 28 at 14:57

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it really such hard work to type out Cauchy-Schwarz in full? Not everybody knows what C-S means!
  $endgroup$
  – TonyK
  Mar 28 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK C-S it's Cauchy-Schwarz inequality.
  $endgroup$
  – Michael Rozenberg
  Mar 28 at 16:18
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Let $u(a^2,b^2)$ and $v(c^2,d^2),$ where $a$, $b$, $c$ and $d$ are positives.

Thus, we need to prove that
$$(a+b)^2+(c+d)^2geqleft(sqrta^2+c^2+sqrtb^2+d^2right)^2$$ or
$$ab+cdgeqsqrt(a^2+c^2)(b^2+d^2),$$ which can be wrong by C-S.

share|cite|improve this answer

answered Mar 28 at 14:57

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it really such hard work to type out Cauchy-Schwarz in full? Not everybody knows what C-S means!
  $endgroup$
  – TonyK
  Mar 28 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK C-S it's Cauchy-Schwarz inequality.
  $endgroup$
  – Michael Rozenberg
  Mar 28 at 16:18
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $u(a^2,b^2)$ and $v(c^2,d^2),$ where $a$, $b$, $c$ and $d$ are positives.

Thus, we need to prove that
$$(a+b)^2+(c+d)^2geqleft(sqrta^2+c^2+sqrtb^2+d^2right)^2$$ or
$$ab+cdgeqsqrt(a^2+c^2)(b^2+d^2),$$ which can be wrong by C-S.

share|cite|improve this answer

answered Mar 28 at 14:57

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it really such hard work to type out Cauchy-Schwarz in full? Not everybody knows what C-S means!
  $endgroup$
  – TonyK
  Mar 28 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK C-S it's Cauchy-Schwarz inequality.
  $endgroup$
  – Michael Rozenberg
  Mar 28 at 16:18
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $u(a^2,b^2)$ and $v(c^2,d^2),$ where $a$, $b$, $c$ and $d$ are positives.

Thus, we need to prove that
$$(a+b)^2+(c+d)^2geqleft(sqrta^2+c^2+sqrtb^2+d^2right)^2$$ or
$$ab+cdgeqsqrt(a^2+c^2)(b^2+d^2),$$ which can be wrong by C-S.

share|cite|improve this answer

answered Mar 28 at 14:57

Michael RozenbergMichael Rozenberg

109k1896201

$endgroup$

share|cite|improve this answer

answered Mar 28 at 14:57

Michael RozenbergMichael Rozenberg

109k1896201

answered Mar 28 at 14:57

Michael RozenbergMichael Rozenberg

109k1896201

answered Mar 28 at 14:57

Michael RozenbergMichael Rozenberg

109k1896201