euclidean metric n square metric?

Question

prove that
p(x,y) <= d(x,y) <= (sqrt(n)) p(x,y) holds in Rn
where
p=square metric define by p(x,y)=max{|x1-y1|, ..., |xn-yn|}
d=euclidean metric define by d(x,y)=sqrt( (x1-y1)^2 + ... + (xn-yn)^2 )
x=(x1,x2,x3,...,xn)
y=(y1,y2,y3,...,yn)

Answer 1

Perhaps this is not rigorous but logically:

|xi-yi| <= p(x,y) for each 1<=i<=n By definition of p.
|xi-yi| ^2 <=p(x,y)^2 {since they are both positive, squaring preserves order}
|x1-y1| ^2 + |x2-y2| ^2 +... |Xn-Yn| ^2 <= p(x,y)^2 + p(x,y)^2 + ...p(x,y)^2 (n times)
|x1-y1| ^2 + |x2-y2| ^2 +... |Xn-Yn| ^2 <= np(x, y)^2.
Now take square roots to get d(x, y) <= (sqrt n) p(,x, y).
Now to show that the Euclidean metric is greater than p(x, y)
We know that if h>=0, then a+h>=a
(adding a positive number to a quantity increases the quantity)
Since p(x,y) = |xi-yi| for some i between 1 and n,
p(x,y) ^2= |xi-yi|^2
and p(x,y) ^2 <= |x1-y1|^2+...|xi-yi|^2 +..|xn-yn|^2 (since we are adding numbers which are greater than or equal to zero)
Now take sqare root to get
p(x, y) <=d(x, y)
Note: |x1-y1|^2 = (x1-y1)^2

Answer 2

We have d(x,y)=sqrt( (x1-y1)^2 + ... + (xn-yn)^2 ) = (x,y)=sqrt( |x1-y1|^2 + ... + |xn-yn|^2 ) . For each i =1,2...n, we have |xi - yi| <= max {|x1-y1|, ..., |xn-yn|} = p(x, y), so that |xi - yi| ^2 <= (p(x,y)^2. Therefore, d(x,y) <= sqrt((p(x,y))^2 ....+ ((p(x,y))^2) (n parcels in the radical). So, d(x,y) <= sqrt(n * (p(x,y))^2) = sqrt(n) * p(x,y), proving the inequality.


Popeye's proof is Ok, it just could be a bit simpler