How I prove this inequality :(a+b+c)(1/a+1/b+1/c)=>9?:(
V Victoria124 New member Joined Oct 18, 2014 Messages 17 Oct 25, 2014 #1 How I prove this inequality a+b+c)(1/a+1/b+1/c)=>9?
pka Elite Member Joined Jan 29, 2005 Messages 11,971 Oct 25, 2014 #2 Victoria124 said: How I prove this inequality a+b+c)(1/a+1/b+1/c)=>9? Click to expand... Write the second factor as \(\displaystyle \dfrac{bc+ac+ab}{abc}\). The carefully multiply the two factors. This is a fact: \(\displaystyle \forall (x~\&~y)[x^2+y^2\ge 2xy]\). You will use that three times.
Victoria124 said: How I prove this inequality a+b+c)(1/a+1/b+1/c)=>9? Click to expand... Write the second factor as \(\displaystyle \dfrac{bc+ac+ab}{abc}\). The carefully multiply the two factors. This is a fact: \(\displaystyle \forall (x~\&~y)[x^2+y^2\ge 2xy]\). You will use that three times.