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
$ab^{2}, ac^{2}, ba^{2}, ca^{2}, cb^{2} $
Since $A.M.\ge G.M.$
$\therefore \frac{ab^{2}+ac^{2}+bc^{2}+ba^{2}+ca^{2}+cb^{2}}{6} $
$\ge\left(ab^{2}. ac^{2}. bc^{2}. ba^{2}. ca^{2}. cb^{2}\right)^{\frac{1}{6}}$
$ \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 $