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I'm having trouble with this:

Use induction to show that $f(x) = \sum\limits_{i=1}^nr_i(x-x_i)^2$ is convex on $\mathbb{R}$ where $\{r_1, \dots, r_n\}$ are positive numbers and $\{x_1, \dots, x_n\}$ are fixed points.

I really don't know where to start or go, any help is appreciated.

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    Prove that a sum of convex functions is convex. – 2017-02-28
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    @dxiv ah, so if I can show that the basis step is convex, then I should know that the sum of all the convex functions would therefore be convex? I guess I keep getting stuck on the argument of showing that $r_1(x-x_1)^2$ is convex. – 2017-02-28
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    You need to show that the base case is convex, indeed, then inductively that a sum of $n+1$ convex functions is convex. For the base case, a positive constant factor $r_1$ doesn't change convexity, so you can just focus on $(x-x_1)^2\,$, which can be proved to be convex either from the definition directly, or using the second derivative sign. – 2017-02-28

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Convexity of $r_1(x-x_1)^2\,$: Notice that the second derivative of this is $2r_1$ which is positive. Hence is strictly convex (reference).

Now induction hypothesis being: $n$ terms summation is convex. Need to prove $n+1$ terms summation is convex. Refer "Prove that the sum of convex functions is again convex." for the proof.