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anonymous
 5 years ago
Use z = a + ib, w = c+id and the triangle inequality to prove that
√[(a+c)² + (b+d)²] ≤ √(a² + b²) + √(c² + d²)
anonymous
 5 years ago
Use z = a + ib, w = c+id and the triangle inequality to prove that √[(a+c)² + (b+d)²] ≤ √(a² + b²) + √(c² + d²)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0square both sides and simplify \[ ac+bd \le \sqrt{a ^{2}+b ^{2}}\sqrt{c ^{2}+d ^{2}}\] square both sides and simplify \[2abcd \le b ^{2}c ^{2} + a ^{2}d ^{2}\] make substitution u =bc v = ad \[2uv \le u ^{2}+v ^{2}\] look at all 3 cases, u=v 2u^2 <= 2u^2 True u>v: u=v+1 \[2v(v+1)\le v ^{2}+(v+1)^{2}\] \[\rightarrow 2v ^{2}+2v \le 2v ^{2}+2v+1\] 0 <= 1 True u<v : u=v1 \[\[2v(v1)\le v ^{2}+(v1)^{2}\] \[\rightarrow 2v ^{2}2v \le 2v ^{2}2v+1\] 0 <= 1 True

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ignore the method above; it is not what they want. Consider z and w. We can consider these as the distances from the origin to each complex number in an argand diagram. Now consider z  w. What does this mean? This is: \[zw = \sqrt{(ac)^2 + (bd)^2} \] so the distance between the two complex numbers, z and w. However, this is not going to be exactly what we want for the LHS, so let us instead consider the complex numbers z and w (and z(w). We work out the values of each of these things: \[ z = \sqrt{a^2 + b^2}\] \[ w = \sqrt{(c)^2 + (d)^2} = \sqrt{c^2 + d^2}\] \[ z(w) = z+w = \sqrt{(a+c)^2 + (b+d)^2}\] But how can we use this to prove the inequality? Wait, DUH, we use the triangle inequality. Considering a triangle with sides a,b,c, we can say: \[a \leq b+c\] with the same holding for each side in a's position. How can this relate to our problem? We know z and w are the distances from the origin to each of these, and z+w is the distance between them. OWAIT these form a triangle. From this we can say: \[z+w \leq z + w \] and the result follows immediately.  Sorry for the long post, the method is actually very quick but as you aren't here for me to guide through it I thought I'd explain everything  this method is FAR better than above (and what they actually wanted) Peace out, INewton.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll be back later to collect my medals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Unfortunately, you gave one to the first post too, so that is unacceptable.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought you were harsh.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you very much! I thought that was a bit harsh too, but the first answer confused me. Yours was very helpful and the answer that I was looking for. Medal given ^^
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