Question

if x+y+z=1 then prove that(1-x)(1-y)(1-z)>8xyz

Answer 1

Hey here is your answer bro:
I have proved the first one like this:

 x+y≥2xy−−√ y+z≥2yz−−√ z+x≥2zx−−√⎫⎭⎬⎪⎪⎪⎪⇒(x+y)(y+z)(z+x)≥8xyz x+y≥2xy y+z≥2yz z+x≥2zx}⇒(x+y)(y+z)(z+x)≥8xyz

But I can't prove or disprove the second one.

Hope its help thank you

Answer 2

Answer:

Step-by-step explanation:

x+ y+z=1

Or, x+y =(1-z) ,y+z=(1-x), x+z=(1-y)

Apply AM >GM for 3 pairs .

(x +y)/2 >√xy

(1-z) >2√xy .

Similarly, (1- y)>2√xz , (1-x)>2√yz

Multiplying 3 equations,

(1-z) (1-x) (1-y) >8xyz