Question

Prove that $6abc<bc(b+c)+ca(c+a)+ab(a+b)$

Answer 1

This will be true only if $a,b,c$ are positive

Apply $AM\ge GM$for $a^{2}b,b^{2}a,c^{2}a,a^{2}c,b^{2}c,c^{2}b$

$⟹\frac{a^{2}b+b^{2}a+c^{2}a+a^{2}c+b^{2}c+c^{2}b}{6}\ge {{\left(\right(abc)}^6)}^{\frac{1}{6}}⟹bc(b+c)+ca(c+a)+ab(a+b)\ge 6abc$

Hence proved