Inequality $|2xy| \leq x^2+y^2$ when determining if function $f(x,y) = xy$ under the constraint $x^4+4xy+y^4=16$

Question

Problem I want to solve is: does the funktion $f(x,y) = xy $ have a global max and/or a global min on the set of values that satisfies $x^4+4xy+y^4=16$

My textbook says: It is enough to prove that the set of values that satisfies $x^4+4xy+y^4=16$ is bounded to some values. To start we have:

$|2xy| \leq x^2+y^2$ which you can see because $(x\pm y)^2 = x^2+y^2 \pm 2xy \geq 0$

And then we get because $(x+y)^2 \geq 0 \iff x^2+2xy+y^2 \geq 0 \iff 4xy \geq -2x^2-2y^2$

That $x^4+4xy+y^4-16 \geq x^4-2x^2-2y^2+y^4-16$. (Thanks for the help with that!)

So far I understand.

But then my textbook states that because $x^4-2x^2-2y^2+y^4-16 = (x^2-1)^2+(y^2-1)^2-18$, it follows that if $x^2>6$ or $y^2 >6$ then $x^4+4xy+y^4-16 > 5^2-18=7 > 0 $

And that means that $x^4+4xy+y^4-16 =0 $ is a bounded set of values. And the set of values are contained in $(x,y): |x| \leq \sqrt{6}, |y| \leq \sqrt{6} $$

I am a little confused. So we are looking to optimize $f$ on the curve $x^4-2x^2-2y^2+y^4-16=0$ We saw that $x^4-2x^2-2y^2+y^4-16 > 7 > 0 $ And because of that the set of values on the curve are bounded?

Answer 1

The problem 1 is $$\left(|x|-|y|\right)^2\geq 0$$ The problem 2: $$(x+y)^2\geq 0$$

Answer 2

For the first problem, you just need to show that $2xy\leq x^2+y^2$ and $-2xy\leq x^2+y^2$. What are these inequalities true?

The second problem becomes (after some simplification): $$2x^2+4xy+2y^2\geq 0.$$

Answer 3

$1$. follows from $(x\pm y)^2=x^2+y^2\pm2xy\ge0$

$2$. subtract $x^4+y^4-16$ from both sides and you'll see it's $4xy\ge-2(x^2+y^2),$

which follows from $1$.

Answer 4

I can write $|2xy|$ as $2xy$ if $2xy\geq0$, or $-2xy$ otherwise. In the first case: $0\leq x^2+y^2-2xy$; in other words $(x-y)^2\geq0$ that is correct $\forall x,y \in R$. In the second case: $0\leq x^2+y^2+2xy$, so: $0\leq (x+y)^2$ that is correct $\forall x,y \in R$.

For the second question: first of all I simplify and obtain: $x^2+y^2+2xy\geq 0$. I can solve as the first case.