If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 &gt; 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
$\dfrac{1 + a + b + c + ab + bc + ac + abc}{7} &gt; \dfrac{a + b + c + ab + bc + ac + abc}{7} \geq \sqrt[7]{a^4b^4c^4}$

where the second inequality follows from the AM-GM inequality.
$(1+a)(1+b)(1+c)=1+(a+b+c)+(ab+bc+ca)+(abc)$
Applying AM≥GM on $(a+b+c)+(ab+bc+ca)+(abc)$
$$\frac{(a+b+c)+(ab+bc+ca)+(abc)}{7}≥(a^{4}b^{4}c^{4})^\frac{1}{7}$$
So $(a+b+c)+(ab+bc+ca)+(abc)≥7(a^{4}b^{4}c^{4})^\frac{1}{7}$
If we add 1 on LHS so now LHS&gt;RHS
So $(1+a)^{7}(1+b)^{7}(1+c)^{7}&gt;7^{7}(a^{4}b^{4}c^{4})$