$a+ b+ c+ d+ e = 12$, and $a^4b^3c^3de=4\times 6^6$, $a,b,c,d,e$ are positive. Find value of $4a + 3b + 3c + d + e$

Question

Given that $$a+ b+ c+ d+ e = 12\quad \&\quad a^4b^3c^3de=4\times 6^6$$ where $a,b,c,d,e$ are positive integers, find the value of $4a + 3b + 3c + d + e$

I tried by trial and error by putting values but it takes to long.

I also tried by taking logarithm on both sides(in equation 2nd) to obtain a similar expression as asked in question but couldn't go any further.

Answer 1

Since nobody wants to type an answer:

as suggested in the comment, use AM-GM inequality on the $12$ numbers $a/4, a/4, a/4, a/4, b/3, b/3, b/3, c/3, c/3, c/3, d, e$.

This gives:

$$\frac{a + b + c + d + e}{12} \geq \sqrt[12]{\frac{a^4b^3c^3de}{4\times 6^6}},$$ where equality holds if and only if all the $12$ numbers are equal.

But the conditions say that both sides are equal to $1$, hence equality does hold, and all the $12$ numbers are equal.

This means that $a/4 = b/3 = c/3 = d = e$. Then from the fact that $a + b + c + d + e = 12$, it follows that $a = 4, b = 3, c = 3, d = 1, e = 1$.