Trig Inequality (partial solutions included)

[lovemake1]

Homework Statement

Given 0 <= a <= b

show that,

a <= sqrt(ab) <= (a+ b / 2) <= b

Homework Equations

a * b <= a^2 / a*b <= a* a

The Attempt at a Solution

I think i know where I am going but i wanted to make sure if its correct so far.

So we know that the order of least to greatest, 0 -> a -> b

and the first part of inequality states that a <= (Sqrt)ab
so i take the sqrt and squre the left side so it makes a ^2.

The inequality is now a^2 <= ab
and this is true because a < b. and a^2 = a * a
there fore a* a is smaller than a * b.

The second part is [sqrt(ab) <= a+ b/ 2]
Sqrt(ab) smaller or equal to (a + b) / 2.

Solution: square both sides, ab <= [ (a+b)/ 2 ]^2
and this gives 4ab <= (a+b)^2
4ab <= a^2 + 2ab + b^2

and last part

a + b / 2 <= b
multiply by 2 to both sides.
a+b <= 2b

and sinec a < b and
a + b <= b * b

is this correct? am i allowed to do what i just did ? please help me

 

[Mark44]

What does this mean -- 0 <\ a <\ b?

Why are the slashes there?

For less than, use <
For less than or equal, use <=
Similar for greater than and greater than or equal.

 

[lovemake1]

Sorry about that, fixed to <=

 

[Mark44]

Can you show that (a + b)/2 is between a and b?
Can you show that sqrt(ab) <= (a + b)/2?

 

[lovemake1]

a) (a+b)/2 between a and b

yes, setting up an equality so that a <= (a+b)/2 <= b
which is same as 2a <= a+b <= 2b

this prooves that (a+b)/2 is between a and b2) sqrt(ab) <= (a+b)/2

square both sides to remove sqrt from ab.

(ab) <= [(a+b)/2]^2
(ab) <= [ (a*b)^2/4]

4ab <= (a*b)^2

is it ok to leave it as 4ab <= (a*b)^2
or should i be expanding so its 4ab <= a^2 + 2ab + b^2 or even go beyond this??

 

[Mark44]

a) (a+b)/2 between a and b

yes, setting up an equality so that a <= (a+b)/2 <= b
which is same as 2a <= a+b <= 2b

this prooves that (a+b)/2 is between a and b

I wouldn't do it that way.
You're given that 0 <= a <= b.
a = (a + a)/2 <= (a + b)/2, since a/2 <= b/2 (which is because a <= b).

It's about as easy to show that (a + b)/2 < b.

2) sqrt(ab) <= (a+b)/2

square both sides to remove sqrt from ab.

(ab) <= [(a+b)/2]^2
(ab) <= [ (a*b)^2/4]

The step above is not obvious (to me). I would expand the (a + b)/2 part.

4ab <= (a*b)^2

is it ok to leave it as 4ab <= (a*b)^2
or should i be expanding so its 4ab <= a^2 + 2ab + b^2 or even go beyond this??

 

[lovemake1]

im really confused, did you just want me to make it 4ab <= a^2 + 2ab + b^2
and subtract 2ab from both sides to make 2ab <= a^2 + b^2 ?

 

[Mark44]

No, subtract 4ab from both sides.

 

[lovemake1]

oh, just make " 0 " on the otherside? (Didnt know this was allowed)
this indefinetly proves that it is greater than left side right?

so i would end up with 0 <= a^2 - 2ab + b^2
making the right side greater.

 

[Mark44]

Sure, you can always add or subtract the same amount from both sides of an equation or inequality.

For 0 <= a^2 - 2ab + b^2, for what values of a and b? That's crucial.

 

[lovemake1]

do i have to find what values a and b have to be ?
a^2 - 2ab + b^2 = (a-b)^2
or just leave it at 0 <= a^2 - 2ab + b^2 to prove sqrt(ab) <= (a+b)/2

from the question it seems like their only asking for the 0 <= a^2 - 2ab + b^2 but also i think i should prove further to see if they are indeed greater than 0.

(a-b)^2 how would this be interpreted as number or equality?
confused.

 

[Mark44]

What you have is a sequence of equivalent inequalities, starting with
sqrt(ab) <= (a + b)/2
and going to 0 <= a^2 - 2ab + b^2

If you know which values the last inequality is true for, you know the values the first one is true for.

 

[lovemake1]

how do i find the values?
im slowly getting lost.

i have to prove that
a^2 - 2ab + b^2 is greater or equal to 0.
but how would i do that?

(a-b)^2 >= 0
any number of A that is smaller than B would give a positive real number.

sry I am lost here

 

[Mark44]

For this problem a - b can be negative or zero, but when you square a - b, you get a number that is always at least zero.

So (a - b)^2 >= 0 for all real numbers a and b. This means that (a + b)/2 >= sqrt(ab) for all nonnegative real numbers a and b. The square root isn't a real number if a and b are opposite in sign, but you're given that both a and b are >= 0.

 

[lovemake1]

ah.. i see the problem.
a * b cannot be a negative number because of the square root.

not sure how it would work out..
ab would always have to > 0

 

[Mark44]

Not a problem. You're given that 0 <= a <= b, so neither one of them can be negative.

 

[lovemake1]

oh right, so it does work out afterall.
0 <= a^2 - 2ab + b^2

this would be the correct final stage then?
given 0 < a < b

 

[Mark44]

No, you don't want to end up with 0 <= a^2 - 2ab + b^2. You want to show from this being true for all values of a and b such that 0 <= a <= b, that sqrt(ab) <= (a + b)/2. That's what you want to conclude in this part of the problem.

Part of doing that is to explain why the inequalities sqrt(ab) <= (a + b)/2 and 0 <= a^2 - 2ab + b^2 have the same solution set, at least for 0 <= a <= b.