Find the minimum of x+y+xy and x+y-xy

[anemone]

Hi MHB,

I've one problem that I think I've already solved half of it, but fact is I really don't know if I am on the right track... that problem is hurting my head so much...

Problem:

For all positive real $x$ and $y$, find the minimum of $x+y+xy$ and $x+y-xy$ if $(x+y+xy)(x+y-xy)=xy$.

Attempt:

AM-GM tells us $\dfrac{(x+y+xy)+(x+y-xy)}{2}\ge \sqrt{(x+y+xy)(x+y-xy)}$ and the LHS simplifies to $x+y$ whereas the RHS simplifies to $\sqrt{xy}$ since $(x+y+xy)(x+y-xy)=xy$. Therefore, we have that $x+y\ge \sqrt{xy}$.

I then naturally think of to subtract $xy$ from both sides of the inequality above and I get

$x+y-xy\ge -xy+\sqrt{xy}$

$x+y-xy\ge -(\sqrt{xy}-0.5)^2+0.25$

Hence, the minimum value for $x+y-xy$ is 0.25.

That's all that I managed to find, and I don't know if this is true, because I don't have the answer to the problem, and I don't know how to proceed to find the minimum value for the other expression and I also don't know how to verify if my first conclusion is correct...

Please, if someone could help me, that would be great and many thanks in advance!

 

[I like Serena]

Hi anemone! (Mmm)

It seems to me you're making it more difficult than it is.

Since $x,y>0$ it follows that $x+y-xy < x+y+xy$.
Therefore the minimum is $x+y-xy$.

Let's call $A=x+y$ and $B=xy$ and note that $A$ and $B$ can take exactly any positive value.
Rewrite and conclude...

 

[anemone]

Hi anemone! (Mmm)

It seems to me you're making it more difficult than it is.

Since $x,y>0$ it follows that $x+y-xy < x+y+xy$.
Therefore the minimum is $x+y-xy$.

Let's call $A=x+y$ and $B=xy$ and note that $A$ and $B$ can take exactly any positive value.
Rewrite and conclude...

Thanks so much for your reply, I like Serena!:)

But, isn't the problem asked us to find for the minimum numerical value for both $x+y-xy$ (or $A-B$) and also, $x+y+xy$ (or $A+B$)?

 

[paulmdrdo1]

From what book did you get this type of problem?

 

[anemone]

From what book did you get this type of problem?

Hi paulmdrdo, this is actually a problem from "The Philippines Mathematical Olympiad" in year 2013.

 

[I like Serena]

Thanks so much for your reply, I like Serena!:)

But, isn't the problem asked us to find for the minimum numerical value for both $x+y-xy$ (or $A-B$) and also, $x+y+xy$ (or $A+B$)?

Ah. It seems I've misinterpreted the way "minimum" was used. (Blush)
I guess the minima should be determined independent of each other then?
In that case the same method still applies.

 

[anemone]

Ah. It seems I've misinterpreted the way "minimum" was used. (Blush)

No worry,I like Serena and thanks for the further reply!

I guess the minima should be determined independent of each other then?

I suppose so...

In that case the same method still applies.

I'm beginning to think I missed something really, seriously important and fundamental here...because I just don't see how to approach it, even after you suggested to think along the line where $x+y+xy>x+y-xy>0$.(Worried)(Sweating)

You know, when we saw how to approach one problem using the method that we have in mind, chances are more likely that we couldn't think out of the box(excuse for covering up my weakness?I guess so.) and if we couldn't see how the suggested solution might be helpful (I'm sure your suggested way is a credible method to solve this particular annoying problem), we've no choice but to stick to our initial plan of attack.

Unfortunately, my initial plan led to nowhere further, for I just found the minimum of $x+y-xy$ to be $\dfrac{1}{4}$.

If chastisement is in order, please go easy on me!:p

 

[I like Serena]

Unfortunately, my initial plan led to nowhere further, for I just found the minimum of $x+y-xy$ to be $\dfrac{1}{4}$.

Zooming in on this for now - before giving away important secrets of the trade. ;)

Suppose we pick $x=y=2$, then $x+y-xy=2+2-2\cdot 2 = 0$...

 

[anemone]

Zooming in on this for now - before giving away important secrets of the trade. ;)

Suppose we pick $x=y=2$, then $x+y-xy=2+2-2\cdot 2 = 0$...

I'm begging you, I like Serena, for telling me the answer(step-by-step preferred) for it, hehehe:p...I'm at my wit end now

but I tried it many more time and here is my ?th attempt:

First, notice that if $x=y=2$, this gives $x+y-xy=0$ and we can rule this possibility out.

Second, if $x=y\ne 2$, all that we have is an equality, $(y+y+y^2)(y+y-y^2)=y^2$, and this only tells us $y=\sqrt{3}=x$

For the case $x>y>0$, since $x+y-xy>0$, we know $x+y>xy$, add the quantity of $xy$ on both sides, we get $x+y+xy>2xy$. Minimum of $x+y+xy$ is obtained if we have the minimum value for $xy$. But we don't have that. So, I go ahead to replace $x+y+xy$ by $\dfrac{xy}{x+y-xy}$ and this resulted in $\dfrac{xy}{x+y-xy}>2xy$, since the quantity of $xy>0$, we divide both sides of the inequality by $xy$ without changing its sign, and get $x+y-xy<\dfrac{1}{2}$. By now, I hope the problem is asking for the maximum value of $x+y-xy$ but not the minimum value of it. Got it, I'll stand in the corner with my nose against the wall for 30 minutes...

 

[Opalg]

For all positive real $x$ and $y$, find the minimum of $x+y+xy$ and $x+y-xy$ if $(x+y+xy)(x+y-xy)=xy$.

If $xy = (x+y+xy)(x+y-xy) = (x+y)^2 - (xy)^2 = x^2 + 2xy + y^2 - (xy)^2$, then $x^2 + y^2 + xy - (xy)^2 = 0.$ Let $k = xy.$ Then $$x^2 + \frac{k^2}{x^2} + k - k^2 = 0,$$ $$x^4 - k(k-1)x^2 + k^2 = 0.$$ That is a quadratic equation in $x^2$, with solutions $$x^2 = \tfrac12\bigl(k(k-1) \pm\sqrt{k^2(k-1)^2 - 4k^2}\bigr) = \tfrac 12k\bigl(k-1 \pm\sqrt{k^2-2k-3}\bigr) = \tfrac 12k\bigl(k-1 \pm\sqrt{(k+1)(k-3)}\bigr).$$ For a start, that tells you that $k\geqslant3$ (otherwise there would be no real solutions for $x^2$). Also, a little experimentation brings up the fact that $$\bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr)^2 = 2k-2 \pm 2\sqrt{(k+1)(k-3)},$$ from which it follows that $x^2 = \frac14k\bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr)^2$ and so $x = \frac12 \sqrt k \bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr).$ Then $$y = \frac kx = \frac{2}{\sqrt k \bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr)}.$$ Rationalise the denominator (I'll omit the details), and you see that $y = \frac12 \sqrt k \bigl(\sqrt{k+1} \mp \sqrt{k-3}\bigr).$ Thus $x+y\pm xy = \sqrt{k(k+1)} \pm k$.

We already know that $k\geqslant3$. When $k=3$, the value of $x+y\pm xy$ is $2\sqrt3 \pm3$. For $k\geqslant3$, both functions $x+y\pm xy$ increase as $k$ increases. Therefore the minimum value of $x+y-xy$ is $2\sqrt3 - 3 \approx0.464$, and the minimum value of $x+y+xy$ is $2\sqrt3 + 3 \approx6.464$, both these these occurring when $x=y=\sqrt3$.

 

[anemone]

If $xy = (x+y+xy)(x+y-xy) = (x+y)^2 - (xy)^2 = x^2 + 2xy + y^2 - (xy)^2$, then $x^2 + y^2 + xy - (xy)^2 = 0.$ Let $k = xy.$ Then $$x^2 + \frac{k^2}{x^2} + k - k^2 = 0,$$ $$x^4 - k(k-1)x^2 + k^2 = 0.$$ That is a quadratic equation in $x^2$, with solutions $$x^2 = \tfrac12\bigl(k(k-1) \pm\sqrt{k^2(k-1)^2 - 4k^2}\bigr) = \tfrac 12k\bigl(k-1 \pm\sqrt{k^2-2k-3}\bigr) = \tfrac 12k\bigl(k-1 \pm\sqrt{(k+1)(k-3)}\bigr).$$ For a start, that tells you that $k\geqslant3$ (otherwise there would be no real solutions for $x^2$). Also, a little experimentation brings up the fact that $$\bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr)^2 = 2k-2 \pm 2\sqrt{(k+1)(k-3)},$$ from which it follows that $x^2 = \frac14k\bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr)^2$ and so $x = \frac12 \sqrt k \bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr).$ Then $$y = \frac kx = \frac{2}{\sqrt k \bigl(\sqrt{k+1} \pm \sqrt{k-3}\bigr)}.$$ Rationalise the denominator (I'll omit the details), and you see that $y = \frac12 \sqrt k \bigl(\sqrt{k+1} \mp \sqrt{k-3}\bigr).$ Thus $x+y\pm xy = \sqrt{k(k+1)} \pm k$.

We already know that $k\geqslant3$. When $k=3$, the value of $x+y\pm xy$ is $2\sqrt3 \pm3$. For $k\geqslant3$, both functions $x+y\pm xy$ increase as $k$ increases. Therefore the minimum value of $x+y-xy$ is $2\sqrt3 - 3 \approx0.464$, and the minimum value of $x+y+xy$ is $2\sqrt3 + 3 \approx6.464$, both these these occurring when $x=y=\sqrt3$.

Thanks so much Opalg for your solution! This problem (I'm sorry to have called it an annoying problem because it isn't, it actually is indeed a brilliant problem!) seems so hard to crack for a person like me and I would still have been stuck with it if you or I like Serena or any helper out there hasn't replied to me. For this I am grateful to both of you. I could never give either you or I like Serena enough thanks, and praises.(Sun)

I believe I just learned something so extremely useful today, and I'm so so happy!(Music) Thanks to MHB too!