Prove that $(1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4$

Question

If a,b,c are positive prove that $(1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4$

My approach I tried (1+a)(1+b)(1+c) =1+[a+b+c+ab+bc+ac+abc] Find the AM of Square Bracket, them AM $\ge$ GM But if I add 1 then AM>GM. Please help me with my approach

Answer 1

You almost did it.

(1+a)(1+b)(1+c) =1+a+b+c+ab+bc+ac+abc (Now subtract 1 and multiply and divide by 7)

= $\frac{7a+7b+7c+7ab+7bc+7ac+7abc}{7}\ge \sqrt[7]{(7)(7a)(7b)(7c)(7ab)(7bc)(7ac)(7abc)}$

add 1 on LHS

=>$1+\frac{7a+7b+7c+7ab+7bc+7ac+7abc}{7} > \sqrt[7]{(7)(7a)(7b)(7c)(7ab)(7bc)(7ac)(7abc)}$

Taking power 7 on both sides

$(1+a)^7(1+b)^7(1+c)^7 > 7^7.a^4 b^4 c^4$