If `a+b+c=1,` then prove that `8/(27a b c)>{1/a-1}{1/b-1}{1/c-1}> 8.`

Explanation

On multiplying both sides by abc, we have to prove that `8//27 gt (1 - a) (1 - b) (1 - c) gt 8 abc`. Now,
`((1 - a) + (a - b) + (1 - c))/(3) gt [(1 - a) (1 - b) (1 - c)]^(1//3)`
`implies (3 - (a + b+ c))/(3) gt [(1 - a)(1 - b)(1 - c)]^(1//3)`
`implies ((2)/(3))^(3) gt (1 -a) (1 - b) (1 - c)` `( :' a + b + c = 1)`
Also, `(a + b)//2) gt (ab)^(1//2), (c + b)//2) gt (cb)`, `(a + c)//2 gt (ac)^(1//2)`
Multiply these three inequanlities, we have
`(a + b)/(2) (b + c)/(2) (a + c)/(2) gt abc`
or `(a + b)(b + c)(c + a) gt 8 abc`
or `(1 - a) (1 - b) (1 - c) gt 8 abc`