Question

For positive real numbers $a,b,c$ such that $a+b+c=p$, which one does not hold?

• A. $\left(p-a\right)\left(p-b\right)\left(p-c\right)\le \frac{8}{27}{p}^3$
• B. $\left(p-a\right)\left(p-b\right)\left(p-c\right)\ge 8abc$
• C. $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\le p$
• D. None of these

Answer 1

Answer: (C)

Using $A.M.\ge G.M.$ one can show that $(b+c)(c+a)(a+b)\ge 8abc$
$\Rightarrow \left(p-a\right)\left(p-b\right)\left(p-c\right)\ge 8abc$
Therefore, $\left(b\right)$ holds. Also,
$\frac{\left(p-a\right)+\left(p-b\right)+\left(p-c\right)}{3}\ge {\left[\left(p-a\right)\left(p-b\right)\left(p-c\right)\right]}^{1/3}$
or $\frac{3p-\left(a+b+c\right)}{3}\ge {\left[\left(p-a\right)\left(p-b\right)\left(p-c\right)\right]}^{1/3}$
or $\frac{2p}{3}\ge {\left[\left(p-a\right)\left(p-b\right)\left(p-c\right)\right]}^{1/3}$
or $\left(p-a\right)\left(p-b\right)\left(p-c\right)\le \frac{8p^3}{27}$
Therefore, $\left(a\right)$ holds. Again,
$\frac{1}{2}\left(\frac{bc}{a}+\frac{ca}{b}\right)\ge \sqrt{\left(\frac{\sqrt{bc}}{a}\frac{ca}{b}\right)}$
and so on. Adding the inequalities, we get
$\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c=p$
Therefore, $\left(c\right)$ does not hold.