Conjecture: Every convex set of area $1$ is contained in a triangle of area $2$. Consider the following:

Conjecture. Every convex set of area $1$ is contained in a triangle of area $A = 2$.

I can prove it if $A$ is changed to $4$.  The convex set being a square shows that $A = 2$ would be tight if true.
It seems like this must be known are at least considered in the literature, but googling hasn't worked for me.  Can anyone supply a proof of this or give me a reference?
 A: Okay, after thinking about it last night I think I have a proof.  Let $a(X)$ be the area of $X$.
Consider a counter-example to the conjecture, with convex set $M$ and minimum enclosing triangle $T$ such that $a(M) < 2 a(T)$.   Without loss of generality, assume that one side of $T$ is contained in the $x$-axis of the plan with its center on $(0, 0)$.  For every point $(x, y)$ in the plane, perform a shear transformation $x \mapsto x + k_1 y$ so that $T$ is isosceles, and then perform a horizontal stretch / compression $x \mapsto k_2 x$ so that $T$ is an equilateral triangle.  Let $T$ and $M$ now refer to these new figures under these transformations.  Since these transformations preserve relative area, take triangles to triangles, take convex sets to convex sets, and are invertable,we know that $T$ and $M$ must still be a counter-example to the conjecture.
So we now have a diagram that looks like this:

Notice I've drawn $T$ intersecting $M$ at the midpoint of each side ($D, E, F$ in the diagram).  This must be if $T$ is a minimal enclosing triangle (see this paper - basically if there was some side that doesn't meet $M$ at the midpoint, you could make a very small rotation of that one side towards the larger part yielding a smaller enclosing triangle).  
Now let's flip $T$ upside-down, create a "star of David" diagram.

Consider $\triangle ADE$.  Notice that $T'$ cuts this triangle into two pieces exactly $1/3$ of the way from the base $DE$ to $A$.  Also notice that $M$ fails to reach the $T'$ line by a distance of $x$.  This is important:  if $M$ did reach this line, than the part of $M$ inside $\triangle ADE$ would be at least $1/3$ of the area of $\triangle ADE$.  If, furthermore, $M$ also reached the line for triangles $\triangle BDF$ and $\triangle CEF$, then $M$ would have area at least
$$
a(M) \geq a(\triangle DEF) + \frac{1}{3}\bigg(a(\triangle ADE) + a(\triangle BDF) + a(\triangle CEF)\bigg) \geq \frac{1}{2} a(T)
$$
This is a clear contradiction.  So, in at least one of the three triangles, $M$ does not reach the $1/3$ mark.   In our example, it fails to reach $T'$ in $\triangle BDF$ and $\triangle ADE$ by a distance of $x$ and $y$ respectively.  However, $M$ exceeds $T'$ by a distance of $z$ in $\triangle CEF$.  We may still have the contradictory statement $a(M) \geq \frac{1}{2} a(T)$ if $z$ is too big compared to $x$ and $y$.   In fact, since we can bound the area of $M$ in each triangle proportional to height, we can say explicity $x + y > z$, since otherwise we could show $a(M) \geq \frac{1}{2} a(T)$.
With the statement $x + y > z$, we can now slide $T'$ without rotating so that it now strictly encloses $M$.  Why?  Well, imagine sliding $T'$ in the direction $\overrightarrow{DF}$.  As $x$ gets smaller, $z$ gets smaller at the same rate, and $y$ stays the same.   Now, if you slide $T'$ in the direction of $\overrightarrow{DE}$, then $y$ will get smaller at the same rate $z$ gets smaller, and since $x + y > z$, we can slide $T'$ so it totally encloses $M$:

This contradicts the fact that $T$ was a minimum triangle (since we can now make $T'$ a bit smaller and still enclose $M$.
Now, one final worry you might have is that there is no absolute minimum enclosing triangle $T$ (perhaps because of an infinite sequence of triangles that get smaller and smaller).   However, do not worry -- any convex set does have an absolute minimum triangle that contains it.  Why?   Well consider the function $F(\alpha, \beta, \gamma)$ that takes in three angles on the range $[0, 2 \pi]$. For angle $\alpha$, we take a line perpendicular to $\alpha$ and approach $M$ (from infinity) with this line until we hit $M$.    If we do this with $\beta$ and $\gamma$, then we create a triangle.  The output of $F(\alpha, \beta, \gamma)$ is the area of this triangle.  While this function may take on infinite values, if we cut these sections out we get a continuous function on a compact domain and therefore must actually achieve it's minimum values.  
