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 $\lambda abc$ , then $\lambda$ is

• A. 2
• B. 1
• C. 6
• D. 3

Answer 1

Answer: (C)

Consider the numbers
$a{b}^2,a{c}^2,b{a}^2,c{a}^2,c{b}^2$
Since $A.M.\ge G.M.$
$\therefore \frac{a{b}^2+a{c}^2+b{c}^2+b{a}^2+c{a}^2+c{b}^2}{6}$
$\ge {\left(a{b}^2.a{c}^2.b{c}^2.b{a}^2.c{a}^2.c{b}^2\right)}^{\frac{1}{6}}$
$\Rightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)$
$\ge 6{\left(a^6{b}^6{c}^6\right)}^{\frac{1}{6}}=6abc$
$\therefore$ Min. value of
$a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)=6abc$
$\therefore \lambda =6$