Question

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 $\lambda abc$, the value of $\lambda$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is D $6$

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