IF a,b >0
Then prove: AM -HM<= (b-a)^2/4a
This is what I did
AM = (a + b)/2
HM = 2/(1/a + 1/b)
So when I subtract AM - HM: I got this
(a+b)/2 - 2ab/(a+b)
[(a+b)^2 - 4ab]/[2(a+b)]
which reduces to [(a-b)^2] / [2(a+b)]
Now what do I do?
IF a,b >0
Then prove: AM -HM<= (b-a)^2/4a
This is what I did
AM = (a + b)/2
HM = 2/(1/a + 1/b)
So when I subtract AM - HM: I got this
(a+b)/2 - 2ab/(a+b)
[(a+b)^2 - 4ab]/[2(a+b)]
which reduces to [(a-b)^2] / [2(a+b)]
Now what do I do?
a and b are the two consecutive terms of a series and they should not be equalOriginally Posted by Nichelle14
I had the choice whether a would be smaller or bigger than b because had not mentioned anything.
I chose a<b because it impies the equation you have to prove.
It is not true.
What about $\displaystyle a=b=2$
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I believe I found a solution.
As my top sentence says, your inequality is false.
It is however true when, for $\displaystyle x,y>0$.
$\displaystyle \phi y\geq x$
Where,
$\displaystyle \phi $ is the solution to the cubic,
$\displaystyle 2x^3+x^2+3x-1=0$.
I don,t think my inequality is false. If you see what we set to prove it includes >= sign. It can be broken into > and = separately. If a=b= any positive real number, AM=GM=HM=a=b which is not of much importance. The real thing lies in inequality part. Hence we have to use a relation between the two terms. However, I don't understand your solution, please explain it.Originally Posted by ThePerfectHacker