Electrical outlet

Retina, 3 bytes

 1

The trailing linefeed is significant.

Input is a space-separated list of unary numbers.

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Explanation

The code simply removes all spaces as well as the 1 after them from the string. Here is why that works:

Addition in unary is simple: just concatenate the numbers which is the same as removing the delimiters. Decrementing by 1 is also simple: just remove a 1 from each number. We want 1 more than the sum of the decremented inputs though, so we simply only remove the 1s we find after spaces, thereby decrementing all but the first input.

Jelly, 3 bytes

’S‘

Decrement (all), sum, increment. Try it here.

Hexagony, 18 14 bytes

.?<_(@'")>{+.!

Unfolded:

  . ? <
 _ ( @ '
" ) > { +
 . ! . .
  . . .

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I don't think side-length 2 is possible, but there must might be a more efficient side-length 3 solution that this.

This is the usual "decrement all, sum, increment" approach, but I'll have to add diagrams later to show how exactly it works in Hexagony.