Question

If $x_1,x_2,...,x_n$ are ${}^{\prime }{n}^{\prime }$ positive real numbers ; then A.M$\ge$G.M$\ge$H.M
$\frac{x_1+x_2+x_3+...+x_n}{n}\ge {(x_1{x}_2{x}_3...x_n)}^n\ge \frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+...+\frac{1}{x_n}}$ equality occurs when numbers are same using this concept.Here equality occurs numbers are of same.Using this concept answer the following

If $a>0,b>0,c>0$ and the minimum value of $a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)$ is $\lambda abc$, then $\lambda$ is

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

Answer: (D)

$6$

As A.M$\ge$G.M$\frac{\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+\frac{a}{b}+\frac{b}{a}}{6}\ge {(\frac{b}{c}\times \frac{c}{b}\times \frac{c}{a}\times \frac{a}{c}\times \frac{a}{b}\times \frac{b}{a})}^{\frac{1}{6}}=1$
$\Rightarrow \frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}+\frac{a^2+b^2}{ab}\ge 6$
$a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge 6abc$
$\therefore$ minimum value of $a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)$ is $6abc$
according to the question, $\lambda abc=6abc$
$\therefore \lambda =6$