Suppose $t$ is a positive integer and suppose $n_i$, $i=1,2,...,2t$ are non-negative integers such that $\sum_{i=1}^{2t}n_i=4t+3$. Minimize $$S={\sum}_{i=1}^{2t}\dfrac{1}{n_i}$$ and determine the optimal choice of integers $n_i$ such that this sum is minimized.
I understand that the equality will be achieved if $n_i$ are more or less equal, so intuitively, the way to ensure that $\sum_i n_i=4t+3$ is $n_i=3$ for $3$ of the $i$'s, and rest all $n_i=2$.
But is there a purely inequality based proof that tells us that this will be the solution?
If two numbers $n_i$ and $n_j$ differ by at least $2$, say, $n_i\ge n_j+2$, then we can make them closer and decrease the sum. Indeed: $$\frac1{n_i}+\frac1{n_j}>\frac1{n_i-1}+\frac1{n_j+1}$$ since $$\frac{n_i+n_j}{n_in_j}>\frac{n_i-1+n_j+1}{(n_i-1)(n_j+1)}$$ since $$n_in_j<(n_i-1)(n_j+1)$$ since $$0<n_i-n_j-1$$ The minimum obviously exists, because there are only finitely many combinations of $n_i$, and if we suppose that it is not when all $n_i$ are equal or differ by $1$, we can make closer the two that differ more and get even lesser sum - a contradiction.
Also there must be $n_i>2$, otherwise the sum of $n_i$ is not greater than $4t$, and must be $n_i<3$ (for $t\ge 2$), otherwise the sum is not less than $6t>4t+3$. Therefore, all $n_i$ are either $2$ or $3$ and there is only one way to make $4t+3$ out of such numbers.
For $t=1$ by the similar reasoning we have $n_1=3,n_2=4$.