Question

If $a,b,c$ be positive and $ab(a+b)+bc(b+c)+ca(c+a)\ge \lambda abc$, then value of $\lambda$ is

Answer 1

For positive quantities $a^{2}b,b{c}^2$,
$A.M.\ge G.M.$
$\frac{a^{2}b+b{c}^2}{2}\ge (a^{2}b.b{c}^2{)}^{\frac{1}{2}}=abc$
$a^{2}b+b{c}^2\ge 2abc$
Similarly,
$b^{2}c+c{a}^2\ge 2abc$ and
$c^{2}a+a{b}^2\ge 2abc$.
Adding the inequalities we get,
$ab(a+b)+bc(b+c)+ac(a+c)\ge 6abc$
Hence, $\lambda =6$