Let $ S \subset \mathbb R^n $ be a convex set.
Given $ \vec x, \vec y, \vec z \in S $ and three positive numbers such that $ a+b+c=1 $, show that $a\vec x+b\vec y+c\vec z$ is in $S$ also.
Ok, so, I have the solution for this, but it doesn't make any sense to me. I'm wondering if someone can produce something more meaningful to me. The only way I know how to prove convexity is the usual $tx + (1-t)y$ approach which this seems to be beyond. Any help here would be appreciated. Thanks!
Here's a hint:
Consider it in two parts. Write $by + cz = (1-a)({b \over {1-a}}y + {c \over {1-a}}z)$, then apply the usual $tX + (1-t)Y$ approach with $Y = by + cz$ and $t = a$.
If you need more of a hint, just give me a shout. :)
In fact, the above pretty much completes it, so let me gives the details.
Consider $tX + (1-t)Y$ with $t = a$, $X = x$ and $Y = by + cz$.
$$ax + by + cz = ax + (1-a)({b \over {1-a}}y + {c \over {1-a}}z) = tX + (1-t)Y \in S$$
since we know that $X = x \in S$ and, by convexity, since $${b \over {1-a}} + {c \over {1-a}} = {{b+c} \over {1-a}} = 1$$ as $a+b+c=1 \iff 1-a = b + c$, we also have $Y = by + cz \in S$.
If this is helpful, then please remember to upvote and/or accept this answer! :)
