3

prove that $f(x,y,z)=\sqrt{\frac{x}{y+2z}}$, $x,y,z>0$ is quasi-convex.

by definition $P_a=\lbrace (x,y,z) : \sqrt{\frac{x}{y+2z}}\geq a\rbrace$ is convex then f is quasi-concave or $P^a=\lbrace (x,y,z) : \sqrt{\frac{x}{y+2z}}\leq a\rbrace$ is convex then f is quasi-convex

let $\textbf{x}=(x,y,z)\in P^a , \textbf{y}=(x_1,y_1,z_1)\in P^a$

$\sqrt{\frac{x}{y+2z}}\leq a$ and $\sqrt{\frac{x_1}{y_1+2z_1}}\leq a$

Let $\lambda \in [0,1]$ and $f(\lambda \textbf{x} + (1-\lambda )\textbf{y})=\sqrt{\frac{\lambda x+(1-\lambda)x_1}{\lambda y+(1-\lambda)y_1+\lambda 2z+(1-\lambda)2z_1}}$

if $\lambda =0$ then $f(\lambda \textbf{x} + (1-\lambda )\textbf{y})=\sqrt{\frac{x_1}{y_1+2z_1}}\leq a$

if $\lambda =1$ then $f(\lambda \textbf{x} + (1-\lambda )\textbf{y})=\sqrt{\frac{x}{y+2z}}\leq a$

if $\lambda \in (0,1)$.

$f(\lambda \textbf{x} + (1-\lambda )\textbf{y})=\sqrt{\frac{\lambda x}{\lambda y+(1-\lambda)y_1+\lambda 2z+(1-\lambda)2z_1}+\frac{(1-\lambda)x_1}{\lambda y+(1-\lambda)y_1+\lambda 2z+(1-\lambda)2z_1}}\leq $

$\leq \sqrt{\frac{\lambda x}{\lambda y+(1-\lambda)y_1+\lambda 2z+(1-\lambda)2z_1}}+\sqrt{\frac{(1-\lambda)x_1}{\lambda y+(1-\lambda)y_1+\lambda 2z+(1-\lambda)2z_1}}\leq$

$\leq \sqrt{\frac{\lambda x}{\lambda y+\lambda 2z}}+\sqrt{\frac{(1-\lambda)x_1}{(1-\lambda)y_1+(1-\lambda)2z_1}}=$

$=\sqrt{\frac{x}{y+2z}}+ \sqrt{\frac{x_1}{y_1+2z_1}}\leq a+a=2a $

and this is the problem, because $2a\geq a$. Can you help me?

Lenin Angarita
  • 33

1 Answers1

1

Actually $f$ is both quasiconvex and quasiconcave, i.e. the function is quasilinear.

Let us denote the domain of $f$ with $$ D = \lbrace (x, y, z) \in \Bbb R^3 \mid x, y, z > 0 \} \rbrace . $$ If $a \le 0$ then $$ P_a=\lbrace (x,y,z) \in D \mid \sqrt{\frac{x}{y+2z}}\ge a\rbrace $$ is equal to $D$. If $a > 0$ then $$ \sqrt{\frac{x}{y+2z}}\ge a \iff \frac{x}{y+2z} \ge a^2\\ \iff x - a^2y -2a^2z \ge 0 $$ so that $P_a$ is the intersection of $D$ with a half-space, and therefore convex.

In the same way one shows that $$ P^a=\lbrace (x,y,z) \in D \mid \sqrt{\frac{x}{y+2z}}\le a\rbrace $$ is either the empty set, or the intersection of $D$ with a half-space, and therefore also convex.

So in any case, both $P_a$ and $P^a$ are convex sets for all $a \in \Bbb R$.

Martin R
  • 113,040