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
it can be proved by using marvelous AM-GM inequality of positive real numbers.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
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.
I think that you have posted a wrong question and I have proved a wrong inequality.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?