# Two Analysis questions

1. Sep 14, 2009

### 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!

2. Sep 14, 2009

### Office_Shredder

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

Can you see how to finish that off?

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