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(Solved): 3. (20 points) (a) Prove that x+yx+y for all real numbers x and y. (b) Use the ...




3. (20 points) (a) Prove that \( |x+y| \leq|x|+|y| \) for all real numbers \( x \) and \( y \).
(b) Use the above inequality
3. (20 points) (a) Prove that for all real numbers and . (b) Use the above inequality to prove that for all real numbers , and . (c) If , prove that .


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In order to solve part a. we will have to first let |x+y| = a, and the and now we will check the conditions for 4 cases:

case 1:

  

let x=1 and y=2

|a|=<=|x|+|y|

put in the values;

  
  
  
Hence for all values of x,y which are greater than or equal to 0 will satisfy the given equation.

Now Case 2:

  

now let x ne -2 and y be -1

so now putting it all in the equation will be:

  
  
  

And now we will have to check case 3:

Where x<=0 and y>=0

so we will let x be -1 and y be 1.

by putting the values in the equation we will get:-

  
  
  

hence this case is also correct

Finally we will check for case 4:

where x>=0 and y<=0

So now we will assume x=1 and y=-1

now by putting the values in the equation we will get:

  
  
  

hence after checking all 4 cases we can conclude that for any real values of x and y the condition given to us is true.
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