Proof that the set $M=\{ (x_1,x_2)\in \mathbb{R}_{++}^2 \mid x_1 x_2\geq k \}$ is convex

Question

$$k\in \mathbb{R}_{+}$$

$$M=\left \{ (x_1,x_2)\in \mathbb{R}_{++}^2 \mid x_1 x_2\geq k\right \}$$

Prove that the set $M$ is convex.

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A hint is given (quoted from the text): We could choose to exploit the fact that if $a$ and $b$ are positive numbers, then: $\frac{a}{b}+\frac{b}{a}\geq 2$

I don't understand how the hint should be used.

This is not an assignment that I have to submit. It's just practice for an upcoming exam.

My understanding is that a set is proven convex if we can take two points from the set and have all the convex combinations of those two points also lie in the set.

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My 1st attempt

I take two vectors from the set $x=(x_1,x_2)\text{ and } y=(y_1,y_2)$

Since the vectors are taken from the set they satisfy the inequality: $x\geq k \text{ and } y\geq k$

Therefore the convex combination of the two must also satisfy the inequality:

$$\forall x,y\in M,\forall k\in\mathbb{R}_{+},\forall \alpha \in [0;1]:\alpha x+(1-\alpha)y\geq k$$

This proves that the set is convex (or does it? I'm not sure at all)

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My 2nd attempt

I simply treat the the inequality $x_1 x_2\geq k$ as an equality $x_1 x_2= k$

I rearrange and get $x_2=\frac{k}{x_1}$

I take the 2nd derivative of $x_{2}$ with respect to $x_1 \Leftrightarrow \frac{d^2 x_2}{dx_1}=\frac{2k}{x_1^3}$ and learn that the it is positive. This proves that the border of the set is stricly convex, which proves that the set is convex (or does it? again not sure).

Answer 1

I disagree with your second attempt, I don't understand it completely though, it's a bit ambiguous what you're saying, but the way I interpret it sounds wrong, but your first attempt is definitely wrong, you have to show that inequality holds for the coordinates of all vectors that lie in the line segment between each two vectors in the set.

Suppose that for some arbitrarily chosen $0\leq \alpha \leq 1$ we write:

$(z_1,z_2)=\alpha (x_1,x_2) + (1-\alpha) (y_1,y_2)$

Does $z_1 \cdot z_2 \geq k$ hold?

Remember that $x_1x_2 \geq k$ and $y_1y_2\geq k$.

Addendum:

This is how I can prove what you want, I don't use $\displaystyle \frac{a}{b} + \frac{b}{a} \geq 2$ in my proof, but since I'm using AM-GM inequality, I guess that with a suitable change of variables you might find a way to write what you want by using that particular inequality (which is implied by AM-GM):

As you said, we have:

$z_1 = \alpha x_1 + (1-\alpha)y_1$

$z_2 = \alpha x_2 + (1-\alpha)y_2$

Just by doing some simple algebraic manipulations, we get:

$z_1 z_2 = \alpha^2 x_1x_2 + \alpha(1-\alpha)x_1y_2 + \alpha(1-\alpha)x_2y_1 + (1-\alpha)^2y_1y_2$

Now use your hypotheses to get:

$z_1z_2 \geq \alpha^2 k + \alpha(1-\alpha)(x_1y_2+x_2y_1) +(1-\alpha)^2k$

But by using AM-GM inequality we have:

$x_1y_2+x_2y_1 \geq 2 \sqrt{x_1x_2}\sqrt{y_1y_2}\geq 2k$

Therefore, we get:

$$z_1z_2 \geq \alpha^2k + \alpha(1-\alpha)(2k)+(1-\alpha)^2k \geq k (\alpha^2+2\alpha(1-\alpha)+(1-\alpha)^2)$$

$$z_1z_2 \geq k (\alpha + (1-\alpha))^2=k$$

Answer 2

The set can be defined by the following linear matrix inequality (LMI)

$$ \begin{bmatrix} x_1 & \sqrt{k} \\ \sqrt{k} & x_2 \end{bmatrix} \succeq 0$$

and, thus, it is a (convex) spectrahedron.