Equations And Inequalities Question 47
Question: For positive real numbers a, b, c such that $ a+b+c=p, $ which one does not hold?
Options:
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
Show Answer
Answer:
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{8p^{3}}{27} $
Therefore, [a] holds, again,
$ \frac{1}{2}( \frac{bc}{a}+\frac{ca}{b} )\ge \sqrt{( \frac{bc}{a}\frac{ca}{b} )} $
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.
BETA
