Reciprocal sum equal to $1$ Given the equation $$1/v + 1/w + 1/x + 1/y + 1/z = 1,$$ where all of $v,w,x,y,z$ are positive integers. One solution is the trivial $1/5 + 1/5 + 1/5 + 1/5 + 1/5$. For distinct solutions with $v&lt;w&lt;x&lt;y&lt;z$ find the minimum of $v+w+x+y+z$.
What is the minimum of $s:=x+y+z+u+v$ subject to $x&lt;y&lt;z&lt;u&lt;v$ and $\frac1x+\frac1y+\frac1z+\frac1u+\frac1v=1$?
From $\frac12+\frac14+\frac17+\frac1{14}+\frac1{28}=1$, we know that $s\le 55$. Hence this is a finite problem.
A quick brute force check of all possibilities with $x+y+z+u+v&lt;55$ reveals that
$$\tag{38} \frac13+\frac14+\frac15+\frac16+\frac1{20}=1$$
with $s=38$ is the optimum.
The next solutions with small sums are:
$$\tag{43} \frac12+\frac14+\frac1{10}+\frac1{12}+\frac1{15}=1 $$
$$\tag{45}  \frac12+\frac14+\frac1{9}+\frac1{12}+\frac1{18}= \frac12+\frac15+\frac1{6}+\frac1{12}+\frac1{20}=1  $$
$$\tag{50} \frac12+\frac14+\frac1{8}+\frac1{12}+\frac1{24}=1 $$
I suspect $2,4,8,12,24$ with sum $50$ will be hard to beat, but don't have a proof (yet).