How prove this $x+y=0$ if $\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$

Question

Question:

let $x,y$ are real numbers,and such

$$\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$$ show that $$x+y=0\tag{1}$$

before I have solve following problem:

if

$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1,\Longrightarrow x+y=0\tag{2}$$ solution:we have $$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{3}$$ $$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{4}$$ $(3)+(4)$ then $x+y=0$

But for $(1)$ we can't use this methods to solve it.maybe can have other nice methods, Thank you

Answer 1

Below some characteristics of the function $f(x,y) = \left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)-y^3$ are shown. The area where $f(x,y)$ is undefined is in gray: $y^2 - x^3 < 0$ or $x^2 + y^3 < 0 $. Therefore the domain of the function is mainly restricted to $x\le 0$ and $y\ge 0$. The line $x+y=0$ is in red, some contour lines of $f(x,y)$ are in blue. The function is somewhat symmetrical around $x+y=0$, but not quite. Shouldn't the restriction $x\le 0$ and $y\ge 0$ be added to the problem in the first place?
It's easy to prove that $x+y=0$ is a solution of $f(x,y)=0$, but proving the reverse seems to be tedious; at least I see no "nice" way to do it.

Answer 2

Rather than use $x$ directly, we will use $z=-x$.

Suppose that $y>0$. Let $$ f_1(z)=z+\sqrt{y^2+z^3} \ (\text{for} \ z\geq{-y^{\frac{2}{3}}}), \ \ f_2(z)=\frac{y^3}{-y+\sqrt{y^3+z^2}} \ (\text{for} \ z\in{\mathbb R}), \ $$

We must then show that $f_1(z) \neq f_2(z)$ when $z\neq y$.

It is easy to see that $f_1$ is increasing on its domain and that $f_2$ is increasing on $(-\infty,0)$ and decreasing on $(0,\infty)$. Let $c=f_1(y)=f_2(y)$.

When $z > y$, we have $f_2(z)<c<f_1(z)$ by the variations of $f_1$ and $f_2$ so $f_1(z) \neq f_2(z)$.

When $0 < z < y$, we have $f_1(z)<c<f_2(z)$ by the variations of $f_1$ and $f_2$ so $f_1(z) \neq f_2(z)$.

When $z <0$, for $f_1(z)$ to be defined we need $z\geq \mu$ where $\mu=-y^{\frac{2}{3}}$. Now $\mu^2=y^{\frac{4}{3}} \leq y^2$, so $f_2(\mu) \geq f_2(y)=c$. As $f_1(z) < c$ by the variations of $f_1$, we still have $f_1(z)<c<f_2(z)$ so $f_1(z) \neq f_2(z)$.

The case $y < 0$ is similar.

Answer 3

We'll replace $x$ on $-x$ and have the following problem.

Let $x$ and $y$ be real numbers such that $$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3.$$ Prove that $$x=y.$$ Indeed, we need to prove that $$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)=\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-x\right)$$ or $$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)-\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)+$$ $$+\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)-\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-x\right)=0$$ or $$\left(\sqrt{y^2+x^3}-\sqrt{x^2+y^3}\right)\left(\sqrt{x^2+y^3}-y\right)+(x-y)\left(\sqrt{x^2+y^3}+x\right)=0,$$ which gives $$x=y$$ or $$\frac{(x^2+xy+y^2-x-y)\left(\sqrt{x^2+y^3}-y\right)}{\sqrt{y^2+x^3}+\sqrt{x^2+y^3}}+\sqrt{x^2+y^3}+x=0,$$ which is $$\sqrt{(x^2+y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2+xy+y^2-y)\sqrt{x^2+y^3}+x^2+xy+y^2-x^2y-xy^2=0.$$ We'll prove that $$\sqrt{(x^2+y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2+xy+y^2-y)\sqrt{x^2+y^3}+x^2+xy+y^2-x^2y-xy^2\geq0,$$ for which we consider four cases.

1. $x\geq0$ and $y\geq0.$

We see that $$\sqrt{(x^2+y^3)(y^2+x^3)}-y\sqrt{x^2+y^3}=\sqrt{x^2+y^3}\left(\sqrt{y^2+x^3}-y\right)\geq0$$ and $$(x^2+xy+y^2)\sqrt{x^2+y^3}-x^2y-xy^2\geq x(x^2+xy+y^2)-x^2y-xy^2=x^3\geq0.$$ The equality occurs for $x=y=0$ only.

2. $x\geq0,$ $y\leq0$.

After replacing $y$ on $-y$ we need to prove that $$\sqrt{(x^2-y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2-xy+y^2+y)\sqrt{x^2-y^3}+x^2-xy+y^2+x^2y-xy^2\geq0,$$ where $x$ and $y$ are non-negatives.

We see that $x^2\geq y^3$, which gives $$x\sqrt{y^2+x^3}\geq x\sqrt{y^2+y^{4.5}}\geq xy^2.$$ The equality occurs only for $x=y=0$ again.

3. $x\leq0$ and $y\geq0.$

We'll replace $x$ on $-x$ and we need to prove that: $$\sqrt{(x^2+y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}+(x^2-xy+y^2-y)\sqrt{x^2+y^3}+x^2-xy+y^2-x^2y+xy^2\geq0,$$ where $x$ and $y$ are non-negatives now.

We see that $$\sqrt{(x^2+y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}=\sqrt{y^2-x^3}\left(\sqrt{x^2+y^3}-x\right)\geq0.$$ Thus, it's enough to prove that $$(x^2-xy+y^2-y)\sqrt{x^2+y^3}+x^2-xy+y^2-x^2y+xy^2\geq0,$$ which is true, but my proof of this statement is still very ugly.

4. $x\le0$ and $y\leq0$.

After replacing $x$ on $-x$ and $y$ on $-y$ we need to prove that:

$$\sqrt{(x^2-y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}+(x^2-xy+y^2+y)\sqrt{x^2-y^3}+x^2-xy+y^2+x^2y+xy^2\geq0,$$ where $x$ and $y$ are non-negative, for which it's enough to prove that $$x^2-xy+y^2\geq x\sqrt{y^2-x^3}.$$ Now, by AM-GM $$x\sqrt{y^2-x^3}\leq\frac{1}{2}(x^2+y^2-x^3).$$ Id est, it's enough to prove that $$x^2-xy+y^2\geq\frac{1}{2}(x^2+y^2-x^3)$$ or $$(x-y)^2+x^3\geq0,$$ which is obvious.

The equality occurs for $x=y=0.$