# Proving convexity of a set

1. Nov 7, 2013

### TaPaKaH

How to show that a set $C=\{(x,y,z)\in\mathbb{R}^3:x\geq0,z\geq0,xz\geq y^2\}$ is convex?

I tried a proof by contradiction: Assume that there exist $c_1=(x_1,y_1,z_1),c_2=(x_2,y_2,z_2)\in C$ and $t\in(0,1)$ such that $tc_1+(1-t)c_2\notin C$.
For this to hold, one would have to have
$$(tx_1+(1-t)x_2)(tz_1+(1-t)z_2)<(ty_1+(1-t)y_2)^2$$ $$t^2x_1z_1+t(1-t)(x_1z_2+x_2z_1)+(1-t)^2x_2z_2<t^2y_1^2+2t(1-t)y_1y_2+(1-t)^2y_2^2.$$ We know that $x_1z_1\geq y_1^2$ and $x_2z_2\geq y_2^2$ but this doesn't seem to help me get any further than $x_1z_2+x_2z_1<2y_1y_2$, from which I can't see a possible contradiction.

2. Nov 7, 2013

### Axiomer

You're on the right track.

Square both sides of $x_1z_2+x_2z_1<2y_1y_2$ to get $x_1^2z_2^2+2(x_1z_1)(x_2z_2)+x_2^2z_1^2<4y_1^2y_2^2$ (*)

By the property of C, $x_1z_1≥y_1^2$ and $x_2z_2≥y_2^2$. Applying this to (*) and rearranging gives $x_1^2z_2^2+x_2^2z_1^2<2y_1^2y_2^2$, but $2y_1^2y_2^2≤2(x_1z_1)(x_2z_2)$, so this gives us $x_1^2z_2^2-2(x_1z_1)(x_2z_2)+x_2^2z_1^2<0$. Why is this a contradiction?

Last edited: Nov 7, 2013
3. Nov 8, 2013

### TaPaKaH

Got it, thank you!