How Can Triangle Inequality Be Applied in Complex Mathematical Proofs?

[BustedBreaks]

1) By writing a = (a+b) + (-b) use the Triangle Inequality to obtain |a| - |b| \leq |a+b|. Then interchange a and b to show that ||a| - |b|| \leq |a+b|.

The replace b by -b to obtain ||a| - |b|| \leq |a - b|.



Okay. I am a bit lost.

I started out by plugging in what they give me for a in the first line into the Triangle Inequality, but that just reduces back to the Triangle Inequality.


I'm just not sure where to start.





2) Let n be a natural number and a_{1}, a_{2}, ... a_{n}be positive numbers. Prove that (1+a_{1})(1+a_{2})+...+(1+a_{n}) \geq 1+a_{1}+a_{2}+...+a_{n}.

and that

(a_{1}+a_{2}+...+a_{n})(a_{1}^{-1}+a_{2}^{-1}+...+a_{n}^{-1}) \geq n^{2})



For the first part of this problem I started out by expanding (1+a_{1})(1+a_{2})+...+(1+a_{n}) for n = 3 and noticed that it would cancel all the terms on the right side making it a bunch of terms greater than or equal to zero, I just couldn't generalize it for n and n+1.

I haven't started on the second part yet.


Thanks!

 

[Office_Shredder]

1) By writing a = (a+b) + (-b) use the Triangle Inequality to obtain |a| - |b| \leq |a+b|. Then interchange a and b to show that ||a| - |b|| \leq |a+b|.

The replace b by -b to obtain ||a| - |b|| \leq |a - b|.
Okay. I am a bit lost.

I started out by plugging in what they give me for a in the first line into the Triangle Inequality, but that just reduces back to the Triangle Inequality. I'm just not sure where to start.

|a+b+(-b)| = |(a-b) + b| <= |a-b| + |b|

Can you see how to finish that off?

2) Let n be a natural number and a_{1}, a_{2}, ... a_{n}be positive numbers. Prove that (1+a_{1})(1+a_{2})+...+(1+a_{n}) \geq 1+a_{1}+a_{2}+...+a_{n}.

and that

(a_{1}+a_{2}+...+a_{n})(a_{1}^{-1}+a_{2}^{-1}+...+a_{n}^{-1}) \geq n^{2})
For the first part of this problem I started out by expanding (1+a_{1})(1+a_{2})+...+(1+a_{n}) for n = 3 and noticed that it would cancel all the terms on the right side making it a bunch of terms greater than or equal to zero, I just couldn't generalize it for n and n+1.

I haven't started on the second part yet.

Do it by induction. If the result holds for n terms, does it hold for n+1? (1+a₁)*...*(a+a_n+1) = (1+a_{2)...(1+an)*1 + (1+a2)...(1+an)*an+1. Use the inductive hypothesis after doing this distribution}