Given vector $x'$ and positive definite matrix $W$, let $$ f(x) := (x-x')^TW(x-x') $$ Show that $f$ is monotonically increasing, i.e. if $y-x' > x-x'$ component-wise, then $f(y) > f(x)$.
I do not know how to approach this. I do not know how to use the positive definiteness of $W$.
On
As showed by Rammus, many more conditions are needed.
In fact, because f is differentiable, for it to be monotonically increasing, its derivative must also be nonnegative, that is:
$\nabla f(x) \ge 0$ for all vector $x$ in $range(f)$.
And this is also a sufficient condition.
So I guess by now you can find the needed conditions for your function $f$.
This is not true if $y \geq x$ is taken to be a componentwise condition -- at least not without further restrictions A simple counterexample is with $x' = 0$, $$ W = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \\~\\ x = (-1, -1), \\~\\ y = (0, 0), $$ then $y \geq x$ and $x^T W x = 2$ but $y^T W y = 0$.
Or in case you are worried about the use of vectors with negative entries in the previous example, here is one in which $x, y \geq 0$. Take $ W = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$, $x = (1, 0)$ and $y = (4/3, 1)$.