If #x, y, z# are positive real numbers such that #x+y+z = 1#. Prove the following inequality: #xy + yz + zx − xyz ≤ 8/27#?

Question

4605 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-23T13:02:35

Please see the proof below

If #x+y+z=1#

Then the maximum occurs when

#x=y=z=1/3#

Therefore,

#xy+yz+zx-xyz=1/3*1/3+1/3*1/3+1/3*1/3-1/3*1/3*1/3#

#=1/9+1/9+1/9-1/27#

#=3/9-1/27#

#=9/27-1/27#

#=8/27#

This is the maximum value,

So,

#AA x,y,z in RR^+# such that #x+y+z=1#

Then,

#xy+yz+zx-xyz<=8/27#