Show that the set $\{(x_1,x_2) : 4 \sqrt{x_1} + x_2 \geq 20\}$ is a convex set. 
Suppose $f : \mathbb{R}^2 \to \mathbb{R}$ is given by $f(x_1,x_2) = 4 \sqrt{x_1} + x_2$. Show that the set $\{(x_1,x_2) : f(x_1,x_2) \geq 20\}$ is a convex set.


We need to show that for any $x = (a,b)$ and $y= (c,d)$ satisfying $\min\{f(x), f(y)\} \geq 20$, $$f(\theta x + (1-\theta)y) \geq 20 \quad \forall\; \theta \in [0,1]$$
How can we show that $4 \sqrt{a} + b \geq 20$ and $4 \sqrt{c} + d \geq 20$ together imply that
$$f(\theta x + (1-\theta)y) = 4 \sqrt{\theta a + (1 - \theta)c} + \theta b + (1-\theta)d \geq 20$$ for all $\theta \in [0,1]$?
 A: The problem boils down to showing: $\sqrt{\theta x+(1-\theta)y}\ge \theta\sqrt{x}+(1-\theta)\sqrt{y}$ as suggested by @Ryszard Szwarc in the comment line above. Apply Cauchy-Schwarz inequality: $\theta\sqrt{x}+(1-\theta)\sqrt{y}=\sqrt{\theta}\sqrt{\theta x}+\sqrt{1-\theta}\sqrt{(1-\theta)y}\le \sqrt{(\theta + (1-\theta))(\theta x+(1-\theta)y)}=\sqrt{\theta x+(1-\theta)y}$. From this you have: $4\sqrt{\theta a+(1-\theta)c}+\theta b + (1-\theta)d \ge 4(\theta\sqrt{a}+(1-\theta)\sqrt{c})+\theta b+(1-\theta)d= \theta(4\sqrt{a}+b)+(1-\theta)(4\sqrt{c}+d)\ge 20\theta+20(1-\theta)=20.$, proving the set convex.
A: The domain mentioned is the supergraph of the function $x_1 \mapsto 20 - 4 \sqrt{x_1}$ from $[0, infty)$ to $\mathbb{R}$. Notice that this function is convex.
Obs: The function $f(x_1, x_2) = 4 \sqrt{x_1} + x_2$ is in fact concave ( sum of two concave functions), and therefore, all its superlevel sets are convex.
Note: One can consider the case of a function that is not concave, yet its superlevel sets are convex. For instance, the function $f\colon (0, \infty)\times \ldots \times (0, \infty) \to (0, \infty)$, $f(x_1, \ldots, x_n) = x_1 \cdots x_n$. Now this function $f$ is quasiconcave. Indeed, say we have two points $(x_1, \ldots, x_n)$, and $(y_1, \ldots, y_n)$ with $\prod x_i$, $\prod y_i\ge a$. We get
$$\prod \frac{x_i+y_i}{2} \ge \prod \sqrt{x_i y_i} \ge \sqrt{a}^2 = a$$. The question is whether the function $f$ is in fact concave. Let's see the case $n=2$. Is it true that
$$\frac{x_1 x_2}{2} + \frac{y_1 y_2}{2} \le \frac{x_1+y_1}{2} \cdot \frac{x_2 + y_2}{2}$$
That depends on the ordering of $x_i$, $y_i$, so it's not always true. Thus, $f$ is quasi-concave, but not concave.
