A) \[(p-a)(p-b)(p-c)\le \frac{8}{27}{{p}^{3}}\]
B) \[(p-a)(p-b)(p-c)\ge 8abc\]
C) \[\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\le p\]
D) None of these
Correct Answer: C
Solution :
| [c] Using \[A.M.\ge G.M.\] one can show that |
| \[(b+c)(c+a)(a+b)\ge 8abc\] |
| \[\Rightarrow \,\,\,(p-a)(p-b)(p-c)\ge 8abc\] |
| Therefore, [b] holds. Also, |
| \[\frac{(p-a)+(p-b)+(p-c)}{3}\ge {{[(p-a)(p-b)(p-c)]}^{1/3}}\] |
| or \[\frac{3p-(a+b+c)}{3}\ge {{[(p-a)(p-b)(p-c)]}^{1/3}}\] |
| or \[\frac{2p}{3}\ge {{[(p-a)(p-b)(p-c)]}^{1/3}}\] |
| or \[(p-a)(p-b)(p-c)\le \frac{8{{p}^{3}}}{27}\] |
| Therefore, [a] holds, again, |
| \[\frac{1}{2}\left( \frac{bc}{a}+\frac{ca}{b} \right)\ge \sqrt{\left( \frac{bc}{a}\frac{ca}{b} \right)}\] |
| And so on. Adding the inequalities, we get |
| \[\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c=p\] |
| Therefore, [c] does not hold. |
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