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Q.
For positive real numbers $a, b, c$ such that $a + b + c = p$, which one does not hold?
Linear Inequalities
Solution:
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.