Algebraic Proof that a Disk is Convex

Question

After searching on Google for a while, I cannot seem to find an algebraic proof that a disk is a convex set. Intuitively, this seems obvious: if you take any two points $x, y$ in a disk, then the line from $x$ to $y$ is clearly contained in the disk.

More specifically, let $D \subset \mathbb{R}^2$ be an origin-centered disk of radius $r$. Let $x = (x_1, x_2), y = (y_1, y_2) \in C$. We wish to show that for any $\lambda \in [0, 1]$, the point $z$ given by $$ z = \lambda x + (1 - \lambda) y = (\lambda x_1 + (1 - \lambda y_1), \lambda x_2 + (1 - \lambda) y_2) $$ also lies in $D$. By definition, we have $\|x\|_2^2 \leq r^2$ and $\|y\|_2^2 \leq r^2$. Using this information, I tried to show that $\|z\|_2^2 \leq r^2$ as follows: \begin{align*} (\lambda x_1 + (1 - \lambda y_1))^2 + (\lambda x_2 + (1 - \lambda) y_2)^2 &= \lambda^2 x_1^2 + 2 \lambda (1 - \lambda) x_1 y_1 + (1 - \lambda)^2 y_1^2 + \\ &\phantom{50} \lambda_2 x_2^2 + 2 \lambda (1 - \lambda) x_2 y_2 + (1 - \lambda)^2 y_2^2 \\ &= \lambda^2 (x_1^2 + x_2^2) + 2 \lambda (1 - \lambda) (x_1 y_1 + x_2 y_2) + (1 - \lambda)^2 (y_1^2 + y_2^2). \end{align*} Unfortunately, I'm stuck at this point -- I can't see a clean way to factor this expression and bound it above by $r^2$. I feel like I am missing obvious, but at this time I cannot determine what that is. Is there a clean proof of this fact that follows from this reasoning?

Edit: Solution using Cauchy-Schwarz Inequality

Even though I used the notation for norms, it didn't occur to me to just use the triangle inequality, as Dustan Levenstein suggested. The solution would then proceed as follows: \begin{align*} \| \lambda x + (1 - \lambda) y \|_2 &\leq \lambda \| x \|_2 + (1 - \lambda) \| y \|_2 \\ &\leq \lambda r + (1 - \lambda) r = r. \end{align*} It follows that $\| z \|_2^2 \leq r^2$, as desired. I'm still going to leave the question up, in case there's a solution that does not rely on the Cauchy-Schwarz inequality.

Answer 1

Equation of a disk: $x^2+y^2≤r^2$ (r is the radius).

Let’s say 2 points in the disk are $(x_1,y_1)$ and $(x_2,y_2)$.

Now we have to show that any point on the line segment joined by these 2 points i.e. $(tx_1+(1-t)x_2, ty_1+(1-t)y_2)$ also satisfies the disk equation, where $1 \leq t \leq 0$.

$(tx_1+(1-t)x_2)^2 + (ty_1+(1-t)y_2)^2$

$= t^2 x_1^2+t^2 y_1^2+(1-t)^2 x_2^2+(1-t)^2 y_2^2 + 2t(1-t)[x_1x_2+y_1y_2]$

$= t^2(x_1^2+y_1^2)+(1-t)^2(x_2^2+y_2^2) + t(1-t)(2x_1x_2 + 2y_1y_2)$

$\leq t^2 r^2+ (1-t)^2 r^2 + t(1-t)(x_1^2+x_2^2+y_1^2+y_2^2)$, because $2xy≤x^2+y^2$,prove it yourself.

$ \leq t^2r^2+ (1-t)^2r^2 + t(1-t)(r^2+r^2) = r^2 [t^2+(1-t)^2 + 2t(1-t)]$

$= r^2$

Hence proven disk is convex.

Answer 2

It's no extra work to prove it for an ellipsoid.

Given a positive (or non negative) definite bilinear form on a denoted $(x,y)$, the convexity of $f(x)=(x,x)$ is, after expanding all terms and cancelling the non-negative scalar factor of $\lambda(1-\lambda)$, the statement that

$$(a,a) + (b,b) \geq (a,b) + (b,a) $$

which is the expansion of $(a-b,a-b) \geq 0$.

I had thought it would be equivalent to Cauchy Schwarz but it looks a little bit weaker.