Question

If $a,b,c\in +R$, such that $a+b+c=p$ then,
$(p-a)(p-b)(p-c)\ge 8abc$

• A. True
• B. False

Answer 1

The correct option is A True

Given $a+b+c=p$

We know that, $A.M.\ge G.M.$

Therefore,

$\frac{(p-a)+(p-b)+(p-c)}{3}\ge \sqrt[3]{3(p-a)(p-b)(p-c)}$

$\frac{3p-(a+b+c)}{3}\ge \sqrt[3]{3(p-a)(p-b)(p-c)}$

$\frac{3p-p}{3}\ge \sqrt[3]{3(p-a)(p-b)(p-c)}$

$\frac{2p}{3}\ge \sqrt[3]{3(p-a)(p-b)(p-c)}$

Cubing on both sides

$(\frac{2p}{3}{)}^3\ge (p-a)(p-b)(p-c)$

$(p-a)(p-b)(p-c)\le \frac{8p^3}{27}$

$\frac{(p-a)(p-b)(p-c)}{8}\le \frac{p^3}{27}$ ————(1)

Similarly,

$\frac{a+b+c}{3}\ge \sqrt[3]{3abc}$

$\frac{p}{3}\ge \sqrt[3]{3abc}$

Cubing on both sides

$\frac{p^3}{27}\ge abc$ ————-(2)

From (1) and (2)

$\frac{(p-a)(p-b)(p-c)}{8}\ge abc$

$(p-a)(p-b)(p-c)\ge 8abc$