Minimize given LCM Find the smallest possible value of $n_1+n_2+\cdots+n_k$ such that $LCM(n_1,n_2,\ldots,n_k)=(2^2)(3^3)(5^5)$. Note that $k$ is not fixed.
I know the answer should be $k=3$, $n_1=2^2$, $n_2=3^3$, and $n_3=5^5$. How do I prove this rigorously?
Note: LCM is least common multiple.
 A: From the AM-Gm inequality you have
$$\frac{n_1+n_2+\dotsb+n_k}{k} \geq \sqrt[k]{n_1n_2\dotsb n_k}.$$
Let $L=\text{lcm}(n_1,n_2, \ldots ,n_k)$ and $G=\gcd(n_1,n_2, \ldots ,n_k)$. Then 
$$n_1n_2\dotsb n_k=LG.$$
Using this into the inequality stated above we get,
$$\frac{S}{k} \geq \sqrt[k]{LG},$$
where $S=n_1+n_2+\dotsb+n_k$.
But $L=2^2 \cdot 3^3 \cdot 5^5$, so to minimizee this we need $G$ to be minimum. The minimum value of GCD of positive integers is $1$, therefore $G=1$. Due to the fact that $L$ is divisible by three primes and $G=1$, we will have $k=3$. Now the answer you got follows.
A: Hint: Clearly $\gcd(n_i, n_j)=1$ for the smallest sum.  Further, for a constant product, the more spread out the numbers are the smaller the sum.  
Hence we are left only with prime powers...
A: If $ k=1$ we have $(n_1+...+n_k)=(n_1)=(2^2.3^3.5^5)$ which is obviously larger than $2^2+3^3+5^5$. If $k>1$, suppose  for some $p\in \{2,3,5 \}$ and for some $i\ne j$ that $p^{A_i}$ divides $n_i$ and $p^{A_j}$ divides $n_j$, with $0 < A_i  \le A_j$. Then we can reduce $n_1+...n_k$ without changing $LCM(n_1,...,n_k)$ by replacing $n_i$ by $n_i/p^{Ai}$. So $n_1+...n_k$ is not minimum unless $n_1,...n_k$ are pairwise co-prime. And obviously we can delete any $n_i$ for which $n_i=1$. This gives $k=3$ and $\{ n_1,n_2,n_3 \} = \{2^2,3^3,5^5 \}$.
