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Can we use hyperbolic function to solve the following problems ?

If $(\sqrt {{y^2-x^3}} - x)(\sqrt {{x^2} + y^3} - y) =y^3$ , prove that $x+ y = 0$

If $(\sqrt {{x^2+y^4}} - x)(\sqrt {{y^2} + x^4} - y) \le x^2 y^2$ , prove that $x + y \ge 0$

EditPiAf
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nam
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  • Do you want see a solution without hyperbolic functions? – Michael Rozenberg Nov 09 '18 at 05:32
  • Yes, of course! Michael Rozenberg, please post your full solutions. – nam Nov 09 '18 at 09:10
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    You need to show us your trying, otherwise this topic will be closed. – Michael Rozenberg Nov 09 '18 at 12:08
  • Sir, I thought $x=\sinh t, y\cosh t$. And nothing more! – nam Nov 10 '18 at 00:08
  • Sir Michael Rozenberg, please! I can't wait because I want to see your solution. – nam Nov 10 '18 at 13:05
  • The first question is a duplicate of https://math.stackexchange.com/questions/533139/ – Andreas Nov 21 '18 at 18:44
  • @MichaelRozenberg Michael, I wonder if there are really "straight" answers to these questions, given that the first question is a duplicate of https://math.stackexchange.com/questions/533139/ and in there, the answer is already pretty intricate. So you might post your answer to shine some light on this. – Andreas Nov 21 '18 at 18:46
  • @Andreas I need to restore my solution, but I am very very busy now. – Michael Rozenberg Nov 21 '18 at 19:29
  • I solved the second problem. The first is indeed, something interesting. Thank you for your interest. – Michael Rozenberg Nov 21 '18 at 20:42
  • @MichaelRozenberg Can you post your solution for the second problem? As the first problem is solved already at math.stackexchange.com/questions/533139 , this will complete the answer. I'd be interested to see how you go about it. Thank you. – Andreas Nov 23 '18 at 08:30
  • @Andreas I posted a solution of the second problem. I think the first problem we can prove by the similar to my proof here: https://math.stackexchange.com/questions/543607 – Michael Rozenberg Nov 26 '18 at 13:19

1 Answers1

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The first problem.

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.

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

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.

  1. $x\leq0$ 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.

  1. $x\le0$ 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 $x=y=0.$

The second problem.

By C-S $$x^2y^2\geq\sqrt{(x^2+y^4)(y^2+x^4)}-x\sqrt{y^2+x^4}-y\sqrt{x^2+y^4}+xy\geq$$ $$\geq |xy|+x^2y^2-x\sqrt{y^2+x^4}-y\sqrt{x^2+y^4}+xy,$$ which gives $$x\sqrt{y^2+x^4}+y\sqrt{x^2+y^4}\geq|xy|+xy.$$ Now, if $xy\geq0$ so since $$x\sqrt{y^2+x^4}+y\sqrt{x^2+y^4}\geq0,$$ we obtain $x\geq0$, $y\geq0$ and from here $x+y\geq0.$

Let $xy<0$, $x>0$ and $y<0$.

Thus, $$x\sqrt{y^2+x^4}\geq-y\sqrt{x^2+y^4}$$ or $$x^2y^2+x^6\geq x^2y^2+y^6$$ or $$x^2\geq y^2,$$ which gives $x+y\geq0$ because $x-y>0.$

Michael Rozenberg
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  • The first part, $x^2y^2\geq\sqrt{(x^2+y^4)(y^2+x^4)}-x\sqrt{y^2+x^4}-y\sqrt{x^2+y^4}+xy$ is what you WANT to establish. The second part, $\sqrt{(x^2+y^4)(y^2+x^4)}-x\sqrt{y^2+x^4}-y\sqrt{x^2+y^4}+xy\geq |xy|+x^2y^2-x\sqrt{y^2+x^4}-y\sqrt{x^2+y^4}$ doesn't hold true in general, e.g. $(x,y)=(1,−1)$ doesn't hold. So I wonder how the proof goes. – Andreas Nov 27 '18 at 20:04
  • @Andreas I work with the given. Your example $(1,-1)$ is not valid. See please better the given. – Michael Rozenberg Nov 27 '18 at 20:16
  • The question doen't say anything about the signs or ranges of $x$ and $y$. – Andreas Nov 27 '18 at 20:18
  • Yes, of course. But I proved that $x+y\geq0$ in any case. I have found a typo. Maybe this is the problem. I fixed. – Michael Rozenberg Nov 27 '18 at 20:20
  • @Andreas We got the last inequality from the given. From the last inequality follows that $x+y\geq0.$ Is the case $xy\geq0$ clear? – Michael Rozenberg Nov 27 '18 at 20:35
  • @Andreas If given that $A\geq B$ for some $x$ and $y$ and we proved that $B\geq C$ for all reals $x$ and $y$ then $A\geq C$ for given $x$ and $y$. – Michael Rozenberg Nov 27 '18 at 20:42
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    Ok I got it. You are of course right that on the condition side, we can make that replacement. – Andreas Nov 27 '18 at 20:44
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    Ok I followed the further steps in your answer and it looks real fine. Congrats! – Andreas Nov 27 '18 at 20:56
  • Dear Mr Micheal Rozenberg, why you replace $x$ become $-x$ in first problem ? – nam Dec 28 '18 at 06:21
  • @nam Because after this substitution we need to prove that $x=y$, which is easier. We can use $x=-a$ and $y=b$. It's the same. – Michael Rozenberg Dec 28 '18 at 06:24
  • Dear Mr Micheal Rozenberg. Please tell me, why you don't consider $xy<0,x<0,y>0$ for second problem ? – nam Dec 28 '18 at 08:05
  • @nam Because the expression is symmetric and it's enough to consider $xy<0$, $x>0$ and $y<0$, which I made. – Michael Rozenberg Dec 28 '18 at 09:21