## 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)

1. anonymous

square 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=v-1 $\[2v(v-1)\le v ^{2}+(v-1)^{2}$ $\rightarrow 2v ^{2}-2v \le 2v ^{2}-2v+1$ 0 <= 1 True

2. anonymous

Ignore 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: $|z-w| = \sqrt{(a-c)^2 + (b-d)^2}$ 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.

3. anonymous

I'll be back later to collect my medals.

4. anonymous

lol

5. anonymous

here

6. anonymous

Unfortunately, you gave one to the first post too, so that is unacceptable.

7. anonymous

I thought you were harsh.

8. anonymous

Thank 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 ^^