How to show $P = \{(x,y)\in \mathbb{R}^2:xy\ge3, x,y\ge0\}$ is convex? Let $P = \{(x,y)\in \mathbb{R}^2:xy\ge3, x,y\ge0\}$, show $P$ is convex.
We have done this kind of problem for hyperplanes, not hyperbolas.
I tried to show it algebraically, by showing that for any convex combination of two points $p_1,p_2\in P$
$$\lambda p_1+(1-\lambda p_2)=(\lambda x_1+(1-\lambda)x_2, \lambda y_1+(1-\lambda)y_2)$$
$$(\lambda x_1+(1-\lambda)x_2)(\lambda y_1+(1-\lambda)y_2)\ge3$$
which would imply that the convex combination is again in $P$.
However, I end up with an enormous sum where nothing cancels elegantly.
Is this even the best approach? If it is, where could I tidy it up?
 A: The conditions $x\ge 0$ and $y\ge 0$ are readily shown.
Assume wlog that $x_1\le x_2$.
If $x_1=x_2$, then the expression for the product simplifies to
$$(\lambda x_1+(1-\lambda)x_2)(\lambda y_1+(1-\lambda)y_2)=x_1(\lambda y_1+(1-\lambda)y_2) =\lambda x_1y_1+(1-\lambda)x_2y_2$$
and so is a convex combination of two numbers $\ge 3$.
In all other cases, the two points are on a line $y=mx+b$. On this line, 
$$ f(x):=xy-3=mx^2+bx-3$$
is a quadratic function of $x$.
If $f(x_1)\ge0$ and $f(x_2)\ge 0$, but $f(x_3)<0$ at some point $x_3$ in between, then $f$ must have two real roots, one root  $\xi_1\in [x_1,x_3)$ and one root $\xi_2\in (x_3,x_2]$. We also conclude that  $m>0$. As $f(0)=-3<0$, we conclude $\xi_1<0<\xi_2$. But then $x_1<0$, contradiction.
A: $(\lambda x_1+(1-\lambda)x_2)(\lambda y_1+(1-\lambda)y_2)\ge 3(\lambda x_1+(1-\lambda)x_2)(\lambda /x_1+(1-\lambda)/x_2 $ because $y_i \ge \frac 1 {x_i}$. Now let $g(x)=\frac 1 x$. $g$ is a convex function on $(0,\infty)$. Hence $g(\lambda / x_1+(1-\lambda)/x_2) \le \lambda x_1+(1-\lambda) x_2$. Now just apply the defintion of $g$ and multiply both sides by $\lambda / x_1+(1-\lambda)/x_2$ to finish the proof. 
