Question

For positive real numbers $a,b$ and $c$ such that $a+b+c=p$, which one holds?

• 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. $Noneoftheabove$

Answer 1

Answer: (A), (B)

The correct options are
A $(p-a)(p-b)(p-c)\le \frac{8}{27}{p}^3$
B $(p-a)(p-b)(p-c)\ge 8abc$

Using $A.M\ge G.M\Rightarrow (b+c)(c+A)(a+b)\ge 8abc$
$\Rightarrow (p-a)(p-b)(p-c)\ge 8abc$
Therefore (b) holds. Also
$\frac{(p-a)(p-b)(p-c)}{3}\ge {\left[\right(p-a\left)\right(p-b\left)\right(p-c\left)\right]}^{1/3}$
$\Rightarrow \frac{3p-(a+b+c)}{3}\ge {\left[\right(p-a\left)\right(p-b\left)\right(p-c\left)\right]}^{1/3}\Rightarrow (p-a)(p-b)(p-c)\le \frac{8p^3}{27}$
Therefore (a) holds. Again
$\frac{1}{2}(\frac{bc}{a}+\frac{ca}{b})\ge \sqrt{\left(\frac{bc}{a}\frac{ca}{b}\right)}$ 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