# Prove that the function $f(x_1 , x_2) = x_1 x_2$ is quasi-concave on $S = \mathbb{R}_{++}^2$

Prove that the function $$f(x_1 , x_2) = x_1 x_2$$ is quasi-concave on $$S = \mathbb{R}_{++}^2$$

I started with the definition of quasi-concavity:

$$f( \lambda x + (1-\lambda) y) \geq \min \left( f(x), f(y) \right)$$

Case 1: $$(λ * x_1 + (1-λ) * y_1) * (λ * x_2 + (1-λ) * y_2) ≥ x_1 * x_2$$ Case 2: $$(λ * x_1 + (1-λ) * y_1) * (λ * x_2 + (1-λ) * y_2) ≥ y_1 * y_2$$

But calculating it for the first case I get: $$x_1 * x_2 * (λ^2-1) + (1-λ) * λ * y_1 * x_2 + (1-λ) * λ * x_1 * y_2 + (1-λ)^2 * y_1 * y_2 ≥ 0$$

And I think now I have to proof that for all positive $$x_1,x_2,y_1,y_2$$ the equation is satisfied. But is that even possible? I think it depends on how one chooses λ.

• I improved a bit. Please improve the rest Jul 12 at 21:42

Let $$(x_1, x_2) \in S$$ and $$(y_1, y_2) \in S$$ and $$c = \min \{ f(x_1, x_2), f(y_1, y_2) \} = \min \{ x_1 x_2, y_1 y_2 \} \, .$$

For $$0 \le \lambda \le 1$$ is $$f(\lambda x_1 + (1-\lambda) y_1, \lambda x_2 + (1-\lambda) y_2) = (\lambda x_1 + (1-\lambda) y_1)(\lambda x_2 + (1-\lambda) y_2) \\ = \lambda^2 x_1 x_2 + \lambda (1-\lambda)(x_1 y_2 + x_2 y_1) + (1-\lambda)^2 y_1 y_2 \, .$$ Now $$x_1 x_2 \ge c$$, $$y_1 y_2 \ge c$$, and using the inequality between arithmetic and geometric mean, $$x_1 y_2 + x_2 y_1 \ge 2 \sqrt{x_1 y_2 x_2 y_1} \ge 2c \, .$$ It follows that $$f(\lambda x_1 + (1-\lambda) y_1, \lambda x_2 + (1-\lambda) y_2) \ge c \bigl( \lambda^2 + 2\lambda(1-\lambda) + (1-\lambda)^2\bigr) = c$$ and that concludes the proof.

Alternatively one can argue that for each $$c > 0$$ the sublevel set $$\{ (x_1, x_2) \in S \mid x_1x_2 \ge c \} = \{ (x,y) \mid x > 0, y \ge \frac c x\}$$ is convex as the epigraph of the convex function $$x \mapsto c/x$$ with the domain $$(0, \infty)$$.