Subset of the $(x,y)$ plane satisfying $x^2 - xy + y^2 \le 0$

Question

The problem:

Describe and Illustrate the region in the $(x,y)$ plane satisfying $x^2 - xy + y^2 \le 0$.

My thoughts:

Write the inequality as $x^2 + y^2 \le xy$.

We can associate with a point (other than the origin) in the plane the right-triangle with hypotenuse its distance from the origin, and legs its projections onto the x and y axes. Then the inequality claims the square of the hypotenuse is less than or equal to the product of the other sides.

This cannot be true for any right triangle since the hypotenuse is greater than either side, so its square must be greater than the product of those sides. Thus the inequality is not satisfied for any point other than the origin.

So the only point where the inequality is satisifed is the origin. Is my reasoning correct?

Can someone please provide me with an algebraic proof so I can be more confident.

Answer 1

$$x^2-xy+y^2=\frac{1}{2}(2x^2+2y^2-2xy)=\frac{1}{2}(x^2+y^2+(x-y)^2)\geq 0$$

with equality iff $$x=y=x-y=0$$

Your reasoning seems fine to me! You just need to be a bit careful where a triangle doesn't exist - i.e. one of the legs has length $0$.

Answer 2

You can transform the inequality into $$ x^2-2xy+y^2\le -xy $$ but also into $$ x^2+2xy+y^2\le 3xy $$ Any solution must therefore satisfy $-xy\ge0$ and $xy\ge0$, because the left hand sides are respectively $(x-y)^2$ and $(x+y)^2$. This forces $xy=0$.

If $x=0$, then you need $y^2\le0$; if $y=0$, then you need $x^2\le0$.

Answer 3

For fun:

AM-GM:

$x^2+y^2 \ge 2|x||y$|, equality for $|x|=|y|$.

Hence:

$2|x||y| \le (x^2+y^2) \le xy$, $x,y$, real,

implies $x=y=0$.