Geometric Triangle Inequality Thm Question!?

Question

Hmm... Okay.

Use the fact that the length of teh vector (x,y) is the square root of x^2 + y^2 and the geometric triangle ineq thm to argue that the square root of a^2 + b^2 + the square root of c^2 + d^2 is greater than the sq root of (a+c)^2 + (b+d)^2 for the vectors (a,b), (c,d) and their sum, (a+c, b+d).

...Get it?

Answer 1

To prove √[(a+c)² + (b+d)²] ≤ √(a+c)² + √(b+d)²

Let X = (a, b), Y = (c, d)
So X + Y = (a + c, b + d)

Triangle inequality says: |X + Y| ≤ |X| + |Y|

Now |X + Y| = √[(a+c)² + (b+d)²]
|X| = √[a² + b²] and |Y| = √[c² + d²]

Thus √[(a+c)² + (b+d)²] ≤ √[a² + b²] + √[c² + d²] QED

Answer 2

dunno your Geometric Triangle Inequality Theorem, but

a^2 +b^2 + c^2 + d^2 +2((a^2 +b^2)(c^2 + d^2))^0.5 >?
(a+c)^2 + (b+d)^2

a^2 + b^2 + c^2 + d^2 +2((a^2 +b^2)(c^2 + d^2))^0.5 >?
a^2 + 2ac + c^2 + b^2 + 2bd + d^2

2((a^2 + b^2)(c^2 + d^2))^0.5 >? 2ac + 2bd

4(a^2 + b^2)(c^2 + d^2) >? (2ac + 2bd)^2

4a^2c^2 +4a^2d^2 + 4b^2c^2 + 4 b^2d^2 >? 4a^2c^2 +4abcd + 4b^2d^2

a^2d^2 + b^2c^2 >? abcd

a^4d^4 + 2a^2b^2c^2d^2 +b^4c^4 > a^2b^2c^2d^2