Sequences and series questions (1 Viewer)

[Jonothancroller]

This is a two part question:
a) Point M divides the interval AB internally in the ratio 1:x in such a way that AM is the geometric mean of BM and BA, find x and draw a diagram.
b) Point M divides the interval externally in the ratio 1:x in such a way that AB is the geometric mean of AM and BM. Find x and draw a diagram.

The second question is also two part:

a) Using the fact that the GM of two numbers cannot exceed their AM, prove that if a, b, c and d are any four positive numbers, (a+b+c+d)/4>(abcd)^1/4 (also the > is suppose to be greater than or equal to)

b) By letting d=(abc)^1/3 in part a), prove that (a+b+c)/3>(abc)^1/3

This one makes intuitive sense, I just don't know where to start with the proof.

I am pretty stuck on this question, thanks for any help.

 

[Jonothancroller]

Bump

 

[psyc1011]

Q2) a)
x = (a+b)/2

y = (c+d)/2

Apply AM-GM definition in question

(x+y)/2 > √(xy)

Sub in x and y,

(a+b+c+d)/4 > √(ab) + √(cd)

Apply AM-GM again with x = √(ab) and y = √(cd)

√(ab) + √(cd) > 2 (abcd)^(1/4) = (abcd)^(1/4) + (abcd)^(1/4) > (abcd)^(1/4)

Nowe have,

(a+b+c+d)/4 > 2 (abcd)^(1/4) = (abcd)^(1/4) as required.

 

[psyc1011]

b)
Let d = (abc)^(1/3) and the RHS becomes

(abcd)^(1/4) = [abc(abc)^(1/3)]^(1/4)
= [(abc)^(4/3)]^(1/4)
= (abc)^(1/3)

So we have

(a+b+c+(abc)^(1/3))/4 > (abc)^(1/3)

Times 4 over

a+b+c+(abc)^(1/3) > 4(abc)^(1/3)

Bring the (abc)^(1/3) over and divide by 3

(a+b+c)/3 > (abc)^(1/3) as requried

 

[Jonothancroller]

Thanks man.

Can anyone help me with this:

a) Point M divides the interval AB internally in the ratio 1:x in such a way that AM is the geometric mean of BM and BA, find x and draw a diagram.
b) Point M divides the interval externally in the ratio 1:x in such a way that AB is the geometric mean of AM and BM. Find x and draw a diagram.

Any help would be appreciated.

 

[psyc1011]

No problem. I think I made the mistake of completely spoon-feeding you the answer.

 

[flavurr]

Thanks man.

Can anyone help me with this:

a) Point M divides the interval AB internally in the ratio 1:x in such a way that AM is the geometric mean of BM and BA, find x and draw a diagram.
b) Point M divides the interval externally in the ratio 1:x in such a way that AB is the geometric mean of AM and BM. Find x and draw a diagram.

Any help would be appreciated.

Is there any other conditions to this question or are any conditions incorrect? (part a)

 

[Jonothancroller]

Is there any other conditions to this question or are any conditions incorrect? (part a)

No, that's the full question. By the way the answer for a) is (-1+root(5))/2, and b) is (3-root(5))/2, I just have no idea how to get there.

 

[rand_althor]

Do something similar for the second part.

 

[Jonothancroller]

Do something similar for the second part.

Thanks man, I could never get b) but that helped a lot.