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Why $f:R^2\to R$, with $\mbox{dom} f=R^2_+$ and $f(x_1,x_2) =x_1x_2$ is quasiconcave?

I have tried to use Jensen eniquality to check that superlevel set $\{x\in R^2_+ | x_1x_2 \ge \alpha\}$ is convex.

$$\begin{align*}(\theta x_1 + (1-\theta) x_3)(\theta x_2 + (1-\theta) x_4) \\= \theta^2 x_1x_2 + (1-\theta)^2x_3x_4+\theta(1-\theta)x_1x_4+\theta(1-\theta)x_2x_3\end{align*}$$ this must be greater or equal to $$\alpha$$

But here i don't know how to prove it.

Yola
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    $x_1x_2 \ge \alpha \iff x_2 \ge {x_1 \over \alpha}$. Use the fact that ${1 \over x}$ is convex in $(0,+\infty)$. – Artur Riazanov Jun 22 '16 at 10:07
  • This is one of those cases where a visual study provides immediate help. Remember, $f(x)$ is quasiconcave if and only if ${x,|,f(x)\geq \alpha}$ is a convex set for any fixed $\alpha$. So pick an $\alpha$, and draw the set---in this case, a single lobe of a hyperbola. It's quite obviously convex. And in fact, if $\alpha\leq 0$, you get the entire positive orthant. I'm not saying that's a proof (which is why this is a comment and not an answer ;-)) but it might very well lead you to one. – Michael Grant Jun 27 '16 at 14:26
  • @MichaelGrant that is generally only useful if one is trying to disprove convexity. generating subsets that are convex does not mean there doesn't exist a subset of points that is not convex. You could try to get counter-examples - but your method generally won't work to prove convexity. – makansij Sep 04 '17 at 19:07
  • As I said, it is not a proof. But if a visualization helps you understand the geometry of the function better, it can help suggest an approach to a proof. For instance, in this case, viewing the set can help you see the functional relationship between $x_1$ and $x_2$. – Michael Grant Sep 04 '17 at 19:21

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Let $(x_1,y_1),\, (x_2,y_2) \in \{x,y\in\mathbb{R}_+: (x,y)\ge \alpha\}$. Then $y_1 \ge {\alpha \over x_1}$ and $y_2 \ge {\alpha \over x_2}$. Since ${1\over x}$ is convex. $${\alpha \over \theta x_1 + (1-\theta)x_2} \le \theta {\alpha \over x_1} + (1-\theta){\alpha \over x_2}$$ Then $${\alpha \over \theta x_1 + (1-\theta)x_2} \le \theta y_1 + (1-\theta)y_2 \implies \alpha \le (\theta x_1 + (1-\theta)x_2)\cdot(\theta y_1 + (1-\theta)y_2) = \theta^2 x_1 y_1 + \theta(1-\theta)(x_2 y_1+x_1 y_2) + (1-\theta)^2 x_2 y_2 = (\theta x_1 + (1-\theta)x_2)(\theta y_1 + (1-\theta)y_2)$$ and that means that $\{x,y\in\mathbb{R}_+: (x,y)\ge \alpha\}$ is convex.

Artur Riazanov
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  • I do not follow the first part: " $(x_1,y_1),, (x_2,y_2) \in {x,y\in\mathbb{R}+: (x,y)\ge \alpha}$ " -> Firstly, the $dom, f = \mathbb{R}{++}^2$, right? So it should be $(x_1,y_1),, (x_2,y_2) \in {x,y\in\mathbb{R}_{++}^2: (x,y)\ge \alpha}$ – makansij Sep 04 '17 at 19:27
  • I'm not seeing how $1 \over x$ helps us if that function has the domain $\mathbb{R}$ and the fucntion we're interested in has the domain $\mathbb{R}_{++}^2$? – makansij Sep 04 '17 at 19:32
  • sorry for the confusion $x$ and $y$ in the expression $x,y \in \mathbb{R}+$ are real numbers, not points. The fact that ${1 \over x}$ is convex gives us the first inequality which leads to the conclusion. The difference between domains is not an issue. The logic behind this formal reasoning is that sets ${x \in \mathbb{R}^2{++} \colon f(x) = a}$ are up to the movement and scaling equal to the graph of ${1 \over x}$. – Artur Riazanov Sep 07 '17 at 08:24
  • I didn't understand your last sentence, sorry. See, this question is related to what you are trying to show: https://math.stackexchange.com/questions/1442241/proving-x-1-x-2-x-1x-2-geq-1-is-convex?rq=1. That's what I'm getting at, I don't understand how we have different domains $\mathbb{R}$ $\mathbb{R^2}$ are necessary for this proof. Or rather - how does it help us? – makansij Sep 08 '17 at 16:09