Question

State true or false:

If $a,b,c\in +R$, such that $a+b+c=p$ then, $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge p$.

• A. True
• B. False

Answer 1

The correct option is A True

Given, $a+b+c=p$

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

$⟹\frac{a+b+c}{3}\ge \sqrt[3]{3abc}$

$⟹\frac{p}{3}\ge \sqrt[3]{3abc}$ —————(1)

Similarly, $\frac{b^2{c}^2+a^2{c}^2+a^2{b}^2}{3}\ge \sqrt[3]{3b^2{c}^2\ast a^2{c}^2\ast a^2{b}^2}$

$⟹\frac{b^2{c}^2+a^2{c}^2+a^2{b}^2}{3}\ge \sqrt[3]{3abc}\ast \sqrt[3]{3(abc{)}^3}$

$⟹\frac{b^2{c}^2+a^2{c}^2+a^2{b}^2}{3abc}\ge \sqrt[3]{3abc}$

$⟹\frac{1}{3}(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c})\ge \sqrt[3]{3abc}$ ———-(2)

From (1) $\sqrt[3]{3abc}$ has a minimum value of $\frac{p}{3}$. Substituting $\sqrt[3]{3abc}=\frac{p}{3}$ we get

$\frac{1}{3}\ast (\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c})\ge \frac{p}{3}$

$(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c})\ge p$