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I want to ask if $$C=\left\{ \left(x,y,z\right)^{T}\mid xyz\geq1,\sqrt{xy}\geq z,x>0,y>0,z>0\right\} $$ is a convex set. I know that:

  • $\forall a\in\left\{ x,y,z\right\} :\left\{ \left(x,y,z\right)^{T}\mid a>0\right\} $ is a half space and therefore a convex set

I want to use a theorem that said that the intersection of convex sets is a convex set,but I don't really whether or not the sets $$\left\{ \left(x,y,z\right)^{T}\mid xyz\geq1\right\} ,\left\{ \left(x,y,z\right)^{T}\mid\sqrt{xy}\geq z\right\} $$ are convex at all. maybe the theorem about intersection is not the best idea to solve this one.

I also tired to proof that $C$ is convex according to the definition of convex sets but it became complicated.

thanks!!

sai-kartik
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Omer Ben
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1 Answers1

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It is a convex set.

Let $(x_1, y_1, z_1)\in C$ and $(x_2, y_2, z_2) \in C$. Let $0\le t \le 1$. We have to prove that $(tx_1 + (1-t)x_2, \ ty_1 + (1-t)y_2, \ tz_1 + (1-t)z_2) \in C$.

First, by Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align} \sqrt{[tx_1 + (1-t)x_2][ty_1 + (1-t)y_2]} &\ge t\sqrt{x_1y_1} + (1-t)\sqrt{x_2y_2}\\ &\ge tz_1 + (1-t)z_2 \end{align} since $\sqrt{x_1y_1}\ge z_1$ and $\sqrt{x_2y_2} \ge z_2$.

Second, by Holder inequality, we have \begin{align} &[tx_1 + (1-t)x_2]\, [ty_1 + (1-t)y_2] \, [ tz_1 + (1-t)z_2]\\ \ge\ & (t\sqrt[3]{x_1y_1z_1} + (1-t)\sqrt[3]{x_2y_2z_2})^3 \\ \ge\ & (t + 1-t)^3 \\ =\ & 1 \end{align} since $x_1y_1z_1 \ge 1$ and $x_2y_2z_2 \ge 1$. Another way: Note that $f(x, y, z) = \sqrt[3]{xyz}$ is concave on $\{(x, y, z) \in \mathbb{R}^3: \ x > 0, y > 0, z > 0\}$. See [How can it be proved that the geometric mean function is concave? Thus, we have \begin{align} &f(tx_1 + (1-t)x_2, \ ty_1 + (1-t)y_2, \ tz_1 + (1-t)z_2)\\ \ge\ & tf(x_1, y_1, z_1) + (1-t)f(x_2, y_2, z_2)\\ \ge\ & 1. \end{align}

We are done.

$\phantom{2}$

Another solution:

Fact 1: Let $m$ and $n$ be positive integers. Denote $x = (x_1, x_2, \cdots, x_n)$. Let $\Omega \subseteq \mathrm{R}^n$ be a convex set. Suppose $f_k(x)$, $k=1, 2, \cdots, m$ are convex functions on $\Omega$. Then, $S = \{x\in \Omega: \ f_k(x) \le 0, \ k= 1, 2, \cdots, m\}$ is convex.

[Proof of Fact 1: Let $u\in S$ and $v\in S$. Let $0 \le t \le 1$. For $k=1, 2, \cdots, m$, we have $f_k(tu+(1-t)v) \le tf_k(u) + (1-t)f_k(v) \le 0$. Thus, $tu + (1-t)v \in S$. We are done.]

Now, for our problem, since $f_1(x, y, z) = -\sqrt[3]{xyz} + 1$ and $f_2(x, y, z) = -\sqrt{xy} + z$ are convex functions on $\{(x, y, z)\in \mathbb{R}^3 : \ x > 0, y > 0, z > 0\}$, by Fact 1, $\{(x, y, z)\in \mathbb{R}^3: \ -\sqrt[3]{xyz} + 1\le 0, \ -\sqrt{xy} + z \le 0, \ x > 0, \ y > 0, \ z > 0\}$ is convex. We are done.

River Li
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  • What did you plug into Cauchy-Bunyakovsky-Schwarz inequality? – Omer Ben Apr 29 '20 at 14:54
  • @OmerBen $(a_1 + a_2)(b_1 + b_2) \ge (\sqrt{a_1b_1} + \sqrt{a_2b_2})^2$, $a_1 = tx_1, a_2 = (1-t)x_2, b_1 = ty_1, b_2 = (1-t)y_2$. – River Li Apr 29 '20 at 14:57
  • Is there any other easier way to do it? I didn't learn about Holder inequality.Thanks! – Omer Ben Apr 29 '20 at 15:16
  • @OmerBen No problem. Actually, Holder inequality is powerful. It might be a good idea to know it. FYI. https://artofproblemsolving.com/wiki/index.php/H%C3%B6lder%27s_Inequality – River Li Apr 29 '20 at 15:26
  • @OmerBen Another way is to use $f(x, y, z) = \sqrt[3]{xyz}$ is concave on ${(x, y, z) \in \mathbb{R}^3: \ x > 0, y > 0, z > 0}$. – River Li Apr 29 '20 at 16:38
  • I show that $f\left(x,y,z\right)=-\sqrt[3]{xyz}$ is convex and then used on the theorem that the sublevel set of $f$ is convex for $t=0$ ,$\Rightarrow$ the set $\left{ \left(\begin{array}{ccc} x & y & z\end{array}\right)^{T}\mid\sqrt{xyz}\geq1\right} $ is convex. but what about the set $\sqrt{xy}\geq z$ ? I can show that for each $t \in \mathbb{R}$ $\sqrt{xy}\geq t$ is convex set but there is no theorem (that I know) about union of convex set. In short :how can I prove that $$\left{ \sqrt{xy}\geq z\mid\left(x,y,z\right)\in\mathbb{R}_{++}^{3},,\sqrt{xyz}\geq1\right} $$ is convex? thanks – Omer Ben May 01 '20 at 12:02
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    @OmerBen I updated my answer, please take a look if it is helpful. – River Li May 01 '20 at 14:27