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 (1+x)(1+y)(1+z)=8 prove xyz≤1

its just application of arithmetic mean>= geometric mean inequality two times

apply A.M. >= G.M. over 

(1+x),(1+y),(1+z)

we have (1+x+1+y+1+z)/3 >= cube root of [(1+x)(1+y)(1+z)]

=> (x+y+z+3 )/3>= cube rot of 8 =2

from this we get x+y+z>= 3..equation 1

again apply A.M.>=G.M. over x,y,z

(x+y+z)/3 >= cube root of (xyz)

putting x+y+z’s least value i.e. 3 from equation 1

3/3>= cube root of (xyz)

=> xyz<= 1

proved