any condition on x andy given????
Q Prove the following
x/y+z +y/x+z +z/x+y≥3/2
x,y,z are all >0
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9 Answers
u can add 3 on both sides ..
then this becomes d A.M > H.M inequality dats d simplest proof !
EDIT : well, even i know juz 2
aage mirka ji
please elaborate a bit more in #3
i am weak in HM inequality
stillwaiting for 24 [3]
lets move on Proof No 2
we have
x2/xz+xy +y2/xy+yz + z2/xz+yz ≥3/2
Applying Cauchy we get
(x+y+z)2/2(x+y+z)≥3/2
x2+y2+z2≥xy+yz+xz
1/2[(x-y)2+(y-z)2+(x-z)2]≥0
hence proved
by symmetry minima if exists is at x=y=z
so put x=y=z u get ans≥3/2
simplest proof -
and if u want the 25 proofs, here it is (from Kaymant sir's page) :
http://kaymant.googlepages.com/HowmanyproofsoftheNesbittsInequanlit.pdf
awesome post mirka
thanx 4 the link
now its party (inequality )time