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 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
 A: We only need to prove that for positive $x$ we have $(1+x)^7\gt 7^{7/3}x^4$. 
For positive $x$, let 
$$f(x)=\frac{(1+x)^7}{x^4}.$$
By using $f'(x)$ we find that $f(x)$ reaches a minimum at $x=4/3$. The minimum value of $f(x)$ is 
$$\frac{7^7}{3^34^4,}$$
which is greater than $7^{7/3}$. 
A: Here we have to prove $$(1+a)(1+b)(1+c)>7(abc)^{\frac{4}{7}}$$
So we get $$1+(a+b+c)+(ab+bc+ca)+abc>7(abc)^{\frac{4}{7}}$$
Now Using $\bf{A.M\geq G.M}\;,$ We get $$a+b+c\geq 3(abc)^{\frac{1}{3}}$$
and $$ab+bc+ca\geq 3(abc)^{\frac{2}{3}}$$
So $$1+\sum a+\sum ab+abc\geq 1+3(abc)^{\frac{1}{3}}+3(abc)^{\frac{2}{3}}+abc$$
Now Again Using $\bf{A.M\geq G.M}\;,$ We get
$$\bf{L.H.S}>1+(abc)^{\frac{1}{3}}+(abc)^{\frac{1}{3}}+(abc)^{\frac{1}{3}}+(abc)^{\frac{2}{3}}+(abc)^{\frac{2}{3}}+(abc)^{\frac{2}{3}}+(abc)\geq 7(abc)^{\frac{4}{7}}$$
A: $\dfrac{1 + a + b + c + ab + bc + ac + abc}{7} > \dfrac{a + b + c + ab + bc + ac + abc}{7} \geq \sqrt[7]{a^4b^4c^4}$
where the second inequality follows from the AM-GM inequality.
A: $(1+a)(1+b)(1+c)=1+(a+b+c)+(ab+bc+ca)+(abc)$
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>RHS
So $(1+a)^{7}(1+b)^{7}(1+c)^{7}>7^{7}(a^{4}b^{4}c^{4})$
