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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 |
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Answer» Correct Answer - 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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