Question

If $a,b,c$ are positive then which of the following holds good?

• A. $b^2{c}^2+c^2{a}^2+a^2{b}^2\ge abc(a+b+c)$
• B. $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c$
• C. $\frac{bc}{a^3}+\frac{ca}{b^3}+\frac{ab}{c^3}\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
• D. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{1}{\sqrt{\left(bc\right)}}+\frac{1}{\sqrt{\left(ca\right)}}+\frac{1}{\sqrt{\left(ab\right)}}$

Answer 1

The correct option is D $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{1}{\sqrt{\left(bc\right)}}+\frac{1}{\sqrt{\left(ca\right)}}+\frac{1}{\sqrt{\left(ab\right)}}$

$\left(A\right)$ AM $\ge$ GM

$\frac{b^2{c}^2+c^2{a}^2+a^2{b}^2}{3}\ge {\left(b^2{c}^2{c}^2{a}^2{a}^2{b}^2\right)}^{1/3}={\left(abc\right)}^{4/3}{b}^2{c}^2+c^2{a}^2+a^2{b}^2\ge 3{\left(abc\right)}^{4/3}\left(1\right)$

Also

$a+b+c\ge 3\left(abc{)}^{1/3}\right(2)$

Dividing $\left(1\right)$ by $\left(2\right)$

$\frac{b^2{c}^2+c^2{a}^2+a^2{b}^2}{a+b+c}\ge \frac{3{\left(abc\right)}^{4/3}}{3{\left(abc\right)}^{1/3}}{b}^2{c}^2+c^2{a}^2+a^2{b}^2\ge \left(abc\right)(a+b+c)$

Hence True

$\left(B\right)$ AM $\ge$ GM

$\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge 3{[\frac{bc}{a}\times \frac{ca}{b}\times \frac{ab}{c}]}^{1/3}=3{\left(abc\right)}^{1/3}\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge 3{\left(abc\right)}^{1/3}\left(1\right)Also,a+b+c\ge 3{\left(abc\right)}^{1/3}\left(2\right)Dividing\left(1\right)by\left(2\right)\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge (a+b+c)$

Hence True

$\left(C\right)$ AM $\ge$ GM

$\frac{bc}{a^3}+\frac{ca}{b^3}+\frac{ab}{c^3}\ge 3{\left[\frac{bc}{a^3}\frac{ca}{b^3}\frac{ab}{c^3}\right]}^{1/3}\frac{bc}{a^3}+\frac{ca}{b^3}+\frac{ab}{c^3}\ge 3{\left[\frac{1}{abc}\right]}^{1/3}\left(1\right)Also,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3{\left[\frac{1}{abc}\right]}^{1/3}\left(2\right)Dividing\left(1\right)by\left(2\right)\frac{bc}{a^3}+\frac{ca}{b^3}+\frac{ab}{c^3}\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Hence True

$\left(D\right)$ AM $\ge$ GM

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3{\left[\frac{1}{abc}\right]}^{1/3}\left(1\right)\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}\ge 3{\left[\frac{1}{abc}\right]}^{1/3}\left(2\right)Dividing\left(1\right)by\left(2\right)\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}$

Hence True