I will prove a related theorem using Cauchy-Schwarz, so that you may attempt to adapt this proof to fit your theorem. If things are still not clear, comment again and I will provide the full details of your proof.
Theorem: Given a point $\vec P$ on a line $L$, every vector $\vec X$ on $L$ has length greater than or equal to the projection of $\vec P$ on a nonzero normal vector to the line.
Note that this shows that the distance between the origin and a point on the line is a minimum when that point is the projection of that point onto the normal vector.
You must set up the problem in a convenient way to apply Cauchy-Schwarz. Start with a point on the line $\vec P$ and a (nonzero) vector normal to the line $\vec N$. Then the line is the set of all points $\vec X=(x,y)$ satisfying $$\vec X\cdot\vec N=\vec P \cdot \vec N$$ (Think about this for a bit, make sure it is consistent with whatever definition of a line you are working with.)
Then by Cauchy-Schwarz we have that $$\lvert \vec P\cdot \vec N\rvert=\lvert \vec X \cdot \vec N\rvert\le\lVert\vec X\rVert\lVert\vec N\rVert$$
And since $\vec N$ is nonzero, we have that $\lVert\vec X\rVert\ge\lvert \vec P \cdot \vec N \rvert/\lVert \vec N \rVert$. Cauchy-Schwarz tells us that equality holds iff $\vec X=t\vec N$ for some scalar $t$, in which case $\vec P \cdot \vec N=\vec X \cdot \vec N = t\vec N \cdot \vec N$ and so $t=\vec P \cdot \vec N / \vec N \cdot \vec N$ and therefore $\vec X $ is the projection of $\vec P$ on $\vec N$.
If you haven't covered Cauchy-Schwarz, and don't want to, you could also treat this using the method of Lagrange's multipliers.