Let ‖ ⋅ ‖ : R n → R be the Euclidean norm. Prove that for any two vectors x , y ∈ R n with x ≠ 0 ≠ y the following inequality holds: | | x ‖ x ‖ − y ‖ y ‖ | | ≤ 2 ‖ x ‖ + ‖ y ‖ ‖ x − y ‖ .I'm very curious as to how I can attack this question. Any hints as well as a full answer is appreciated. I've tried using the triangle inequality (which is, basically, all I could think about) with no luck. I guess I could try using the definition of the euclidean norm letting x = ( x 1 , … , x n ) and y = ( y 1 , … , y n ) and try something, but I don't want to go down that path (and I don't think it would lead me anywhere, either).

Question

Let   ‖  ⋅  ‖  :            R        n    →      R   be the Euclidean norm. Prove that for any two vectors   x  ,  y  ∈            R        n   with   x  ≠  0  ≠  y the following inequality holds:      |          |              x                  ‖          x          ‖                    −              y                  ‖          y          ‖                    |        |    ≤      2          ‖      x      ‖      +      ‖      y      ‖        ‖  x  −  y  ‖  .I&#039;m very curious as to how I can attack this question. Any hints as well as a full answer is appreciated. I&#039;ve tried using the triangle inequality (which is, basically, all I could think about) with no luck. I guess I could try using the definition of the euclidean norm letting   x  =  (      x    1    ,  …  ,      x    n    ) and   y  =  (      y    1    ,  …  ,      y    n    ) and try something, but I don&#039;t want to go down that path (and I don&#039;t think it would lead me anywhere, either).

Haylee Branch · Accepted Answer

Let x=ru and y=sv where u and v are unit vectors,   ‖  x  ‖  =  r and   ‖  y  ‖  =  s. Then your inequality says  ‖  u  −  v  ‖  ≤  2            ‖      r      u      −      s      v      ‖              r      +      s           for     r  ,  s  &amp;gt;  0By homogeneity, we may assume r+s=1. Thus the inequality becomes  ‖  u  −  v  ‖  ≤  2  ‖  r  u  −  (  1  −  r  )  v  ‖     for     0  &amp;lt;  r  &amp;lt;  1The right side is convex as a function of r. Now there is a linear isometry that interchanges u and v, so   ‖  r  u  −  (  1  −  r  )  v  ‖  =  ‖  (  1  −  r  )  u  −  r  v  ‖Thus the minimum must occur at   r  =  1      /    2

Dana Russo · Accepted Answer

Let x=ur and y=sv where   r  ,  s  ∈            R        +   and   ‖  u  ‖  =  ‖  v  ‖  =  1.Let c=s/r. Then c&amp;gt;0.The LHS of the inequality is   ‖  u  −  v  ‖  .. The RHS is       2          1      +      c        ‖  u  −  c  v  ‖  .We have   ‖  u      ‖    2    =  ‖  v      ‖    2    =  1.. And 1+c&amp;gt;0. So we have  ‖  u  −  v  ‖    ≤        2          1      +      c        ‖  u  −  c  v  ‖    ⟺    (  1  +  c  )  ⋅  ‖  u  −  v  ‖    ≤    2  ‖  u  −  c  v  ‖    ⟺    (  1  +  c      )    2    ⋅  ‖  u  −  v      ‖    2      ≤    4  ‖  u  −  c  v      ‖    2      ⟺    (  1  +  c      )    2    ⋅  (  ‖  u      ‖    2    +  ‖  v      ‖    2    −  2  u  ⋅  v  )    ≤    4  ⋅  (    ‖  u      ‖    2    +      c    2    ‖  v      ‖    2    −  2  c  (  u  ⋅  v  )    )    ⟺    (  1  +  c      )    2    ⋅  (  2  −  2  u  ⋅  v  )  ≤  4  (  1  +      c    2    −  2  c  (  u  ⋅  v  )  )    ⟺    (  −  2  u  ⋅  v  )  ⋅  (    (  1  +  c      )    2    −  4  c  )    ≤    4  (  1  +      c    2    )  −  2  (  1  +  c      )    2      ⟺    (  −  2  u  ⋅  v  )  (  1  −  c      )    2    ≤  2  (  1  −  c      )    2    .Now       |    −  2  u  ⋅  v      |    ≤  2  ‖  u  ‖  ⋅  ‖  v  ‖  =  2.. So the absolute value of the LHS of the last line above is at most   2  (  1  −  c      )    2    .

Let ∥*∥:RR^n to RR be the Euclidean norm. Prove that for any two vectors x,y in RR^n with x != 0 != y the following inequality holds: norm( (x)/(norm(x))-(y)/(norm(y))) <=(2)/(norm(x)+norm(y))norm(x-y)

Answered question

Answer & Explanation

New Questions in Precalculus