Proving l a + b l < l 1 + ab l with l a l < 1 and l b l < 1

  • Thread starter Andrax
  • Start date
In summary, to prove the given identity, it is necessary to show that the inequality |a + b| < 1 + |ab| holds true for all values of a and b. This can be done by using the triangle identity for absolute values and multiplying both sides of the inequality by |ab|, as long as a and b are both non-zero. However, if either a or b is equal to 0, then the identity is automatically satisfied.
  • #1
Andrax
117
0

Homework Statement


so this is the question I've trid my best (over 2 hours trynig this and failed) can someone please help me
show that :
l a l < 1 and l b l<1 => l a + b l < l 1 + ab l

Homework Equations



just a little bit on info , you can only stay from the lal < 1 and lbl <1 side to get to the other one

The Attempt at a Solution


here's what i tried so fay
l a l < 1 and l b l<1 => l a b l < 1
=> -1 < ab < 1
=> 0 < 1 + ab <2
=> l 1 + ab l <2
l a l < 1 and l b l<1 => -2 < a + b < 2
=> l a + b l < 2
yup I am pretty much stuck here..pleasee help(i can't say l a + b l < l 1 + ab l because they are both<2..
 
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  • #2
Andrax said:

Homework Statement


so this is the question I've trid my best (over 2 hours trynig this and failed) can someone please help me
show that :
l a l < 1 and l b l<1 => l a + b l < l 1 + ab l

Homework Equations



just a little bit on info , you can only stay from the lal < 1 and lbl <1 side to get to the other one

The Attempt at a Solution


here's what i tried so fay
l a l < 1 and l b l<1 => l a b l < 1
=> -1 < ab < 1
=> 0 < 1 + ab <2
=> l 1 + ab l <2
l a l < 1 and l b l<1 => -2 < a + b < 2
=> l a + b l < 2
yup I am pretty much stuck here..pleasee help(i can't say l a + b l < l 1 + ab l because they are both<2..

Don't worry too much about staying on one side while you figure out what's going on. You can always modify your argument to fix that later. You can even work backwards while analyzing it. So far you have 0 < 1+ab, so your problem is equivalent to |a+b|<1+ab. Write that as$$
\sqrt{(a+b)^2} < 1+ab$$Square both sides and work on that. Once you see it, modify it by working backwards through the reversible steps.
 
  • #3
if |a|=0.8 which is less than 1 and |b|=0.2 which is also less than 1 but |ab|= 1.6 which is greater than 1. so your very first line of the solution is wrong
 
  • #4
|a|+|b| ? |a + b|
|a||b| ? |ab|
As LCKurtz says - don't worry about the rules while you are trying to understand the relations.

@KUMAR |0.8x0.2| = 0.16 < 1 isn't it?
 
  • #5
kumar sidhant said:
if |a|=0.8 which is less than 1 and |b|=0.2 which is also less than 1 but |ab|= 1.6 which is greater than 1. So your very first line of the solution is wrong

##(.2)( .8) = .16##
 
  • #6
LCKurtz said:
Don't worry too much about staying on one side while you figure out what's going on. You can always modify your argument to fix that later. You can even work backwards while analyzing it. So far you have 0 < 1+ab, so your problem is equivalent to |a+b|<1+ab. Write that as$$
\sqrt{(a+b)^2} < 1+ab$$Square both sides and work on that. Once you see it, modify it by working backwards through the reversible steps.

thank you
 
  • #7
tried this solution?

let us assume that
|a+b| < | 1+ab|
squaring both sides
a*a + 2ab + b*b < 1 + 2ab + a*a*b*b ----- 1

we have
|a |< |1|
square both side
a*a < 1
multiply with b*b it is positive and no change in inequality
a*a*b*b < b*b
adding 1 on both sides and 2ab
1 + 2ab +a*a*b*b < 1 + 2ab + b*b ---- 2
using 1 and 2
we have
a*a + 2ab + b*b < 1+ 2ab + b*b
which boil down to
a*a < 1 which is true
I do not know whether this approach is right or wrong
 
  • #8
Andrax said:
tried this solution?

let us assume that
|a+b| < | 1+ab|
squaring both sides
a*a + 2ab + b*b < 1 + 2ab + a*a*b*b ----- 1

we have
|a |< |1|
square both side
a*a < 1
multiply with b*b it is positive and no change in inequality
a*a*b*b < b*b
adding 1 on both sides and 2ab
1 + 2ab +a*a*b*b < 1 + 2ab + b*b ---- 2
using 1 and 2
we have
a*a + 2ab + b*b < 1+ 2ab + b*b
which boil down to
a*a < 1 which is true
I do not know whether this approach is right or wrong

I would call that an exploratory argument. It shows you why the inequality works so you understand it. But to make a correct argument, you must start with something that is true and end up with your inequality. So your argument would start like this:

|a| < 1 is given
|a|2 = a2 < 1 ...

Continue by reversing your steps and ending up with what you were to prove.
 
  • #9
LCKurtz said:
I would call that an exploratory argument. It shows you why the inequality works so you understand it. But to make a correct argument, you must start with something that is true and end up with your inequality. So your argument would start like this:

|a| < 1 is given
|a|2 = a2 < 1 ...

Continue by reversing your steps and ending up with what you were to prove.

thanks again i think i got it :)
 
  • #10
Hint:
If [itex]a = 0[/itex] or [itex]b = 0[/itex], then the identity is surely satisfied (why?).

If [itex]a \neq 0[/itex] and [itex]b \neq 0[/itex], then [itex]\vert a \, b \vert > 0[/itex]:
Apply the triangle identity for the absolute value to:
[tex]
\left\vert \frac{1}{a} + \frac{1}{b} \right\vert
[/tex]
and multiply both sides of the inequality by [itex]\vert a \, b \vert[/itex].
 

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