This isn't a proof. It's an illustration of what is happening. Imagine that we can bear-in a checker exactly or play it in the outfield. (Ignore bearing in while wasting a pip for the moment.) The number of pips remaining in the outfield will be identical with either play. However, there will be one fewer checker in the outfield after bringing one home to the six point.
I've posted four positions below. They are the four possible positions with exactly seven pips left to bear-in (with one or two checkers left.) The position with one checker will always be at least as good and often better than any of the other positions. When you roll seven or more pips, you will always get your last checker home when you only have to move one checker. When you have to move two checkers, you only bring them both in with one of the three possible sevens. The inefficiency of two checkers partially extends to rolls of eight or more. You always prefer one checker left to two checkers left when the pip count is identical.
This abstraction extends to other positions with more checkers. You are always better with one fewer checker in the outfield (except for problem positions mentioned in my other post) when the pip count is identical. The difference in any two positions is that after bearing in one checker exactly, you have made a substitution of two checkers for one similar to the illustration below. One checker is always more efficient than two. Two is more efficient than three, etc. Therefore, always bear-in to the six with one number when you can do so exactly.
(Yes, I know that you need doubles to get off in the positions shown. Please ignore that and focus on the question of an efficient bear-in with a seven.)
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| Board image courtesy of GO-Figure | |
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| Board image courtesy of GO-Figure | |
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| Board image courtesy of GO-Figure | |
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| Board image courtesy of GO-Figure | |