I need to prove:

for a,b,c > 0

prove a/(b+c) + b/(a+c) + c/(b+a) is less than or equal to 3/2

Printable View

- Jun 3rd 2006, 08:05 PMNichelle14Proof
I need to prove:

for a,b,c > 0

prove a/(b+c) + b/(a+c) + c/(b+a) is less than or equal to 3/2 - Jun 4th 2006, 08:19 AMmalaygoelQuote:

Originally Posted by**Nichelle14**

let x=a/(b+c) + b/(a+c) + c/(b+a)

now, b+c>=2(bc)^(1/2)

hence,1/(b+c)<=1/2(bc)^(1/2)

hence,2x<=a/(bc)^(1/2) + b/(ac)^(1/2) + c/(ba)^(1/2)

again applying AM-GM inequality on the right hand side of the equation

2x<=3

x<=3/2 - Jun 4th 2006, 09:42 AMrgep
AM/GM is your friend. Here's another way, a bit longer but with a couple of useful tricks.

Firstly observe that = , using AM/GM twice.

In your case this says that . I claim that this last sum is . We can divide through by c and then we need to prove that . Fix b and treat as a function of a. It is minimised when the derivative = 0 and this is at and the value is . A further application of AM/GM shows that whatever F is, so the minimum is always , as claimed. - Jun 19th 2006, 09:32 PMmalaygoelQuote:

Originally Posted by**Nichelle14**

Let a=1, b=2 ,c=3

then

I have started a new thread 'Proof on inequality'in "pre algebra and algebra" asking for proof of the inequality .

Could someone please help how I proved an inequality which does not exist?