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For positive real numbers `a ,bc` such that `a+b+c=p ,` which one holds? `(p-a)(p-b)(p-c)lt=8/(27)p^3` `(p-a)(p-b)(p-c)geq8a b c` `(b c)/a+(c a)/b+(a b)/clt=p` `non eoft h e s e`

A

`(p-a)(p-b)(p-c)le(8)/(27)p^3`

B

`(p-a)(p-b)(p-c)gt 8abc`

C

`(bc)/(a)+(ca)/(b)+(ab)/(c)le p`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A, B
Using A.M `ge` G.M one can show that
`(b + c) (c + a) (a + b) ge 8abc`
`implies (p - a) (p - b) (p -c ) ge 8abc`
Therefore (2), hold. Also
`((p -a) + (p - b) + (p -c))/(3) ge [(p - a)(p - b) (p - c)]^(1//3)`
or `(3p - (a + b + c))/(3) ge [(p - a)(p - b)(p - c)]^(1//3)`
or `(2p)/(3) ge [(p - a)(p - b)(p - c)]^(1//3)`
or `(p -a) (p - b) (p - c) le (8p^(2))/(27)`
Therefore (1) holds. Again
`(1)/(2) ((bc)/(a) + (ca)/(a)) ge sqrt(((bc)/(a) (ca)/(a)))`
and so on. Adding the inequalities, we get
`(bc)/(a) + (ca)/(b) + (ab)/(c ) ge a + b + c = p`
Therefore, (3) does not hold.
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