nguoivn wrote:
Given a,b,c>0. Prove that: (1+a+b+c)(1+ab+bc+ca)≥4√2√(a+bc)(b+ca)(c+ab)
Note that
(1+a+b+c)(1+ab+bc+ca)=(1+a)(1+b)(1+c)+(a+b)(b+c)(c+a)≥2√(1+a)(1+b)(1+c)(a+b)(b+c)(c+a).
Therefore, it suffices to prove that
8(a+bc)(b+ca)(c+ab)≤(1+a)(1+b)(1+c)(a+b)(b+c)(c+a).
By the AM-GM Inequality, we have
2√(a+bc)(b+ca)≤a+bc+b+ca=(c+1)(a+b),
2√(b+ca)(c+ab)≤(a+1)(b+c),
2√(c+ab)(a+bc)≤(b+1)(c+a).
Multiplying these inequalities, we get the result.