Question

If a, b, c be positive, prove that $(b+c)(c+a)(c+b)\ge 8abc$

Answer 1

Since the arithmetic mean of two positive numbers cannot be less than their geometric mean, therefore
$\frac{b+c}{2}\ge \sqrt{bc}............\left(i\right)$
$\frac{c+a}{2}\ge \sqrt{ca}............\left(i\right)$
$\frac{c+a}{2}\ge \sqrt{ca}............\left(i\right)$
Multiplying the corresponding sides of the above inequalities , we obtain,
$\frac{1}{8}(b+c)(c+a)(a+b)\ge abc$,
i.e.,
$(b+c)(c+a)(c+b)\ge 8abc$