symmetric algebra

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Bluesoul

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Bluesoul

#1 h Mar 12, 2022, 4:29 AM • 1 Y

Y by ImSh95

If $ab+\sqrt{ab+1}+\sqrt{a^2+b}\sqrt{a+b^2}=0$ , find the value of $b\sqrt{a^2+b}+a\sqrt{b^2+a}$

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MortemEtInteritum

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MortemEtInteritum

#2 h Mar 12, 2022, 4:33 AM • 1 Y

Y by ImSh95

The first equation becomes $(ab+\sqrt{a^2+b}\sqrt{a+b^2})^2=(-\sqrt{ab+1})^2$ . This rearranges to $a^3+2a^2b^2+b^3+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$ . Now let $x=b\sqrt{a^2+b}+a\sqrt{b^2+a}$ ; then we get $x^2=a^3+2a^2b^3+b^2+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$ , so $x=\pm 1$ . Now suppose $x=-1$ . Note that $ab\leq 0$ , since $\sqrt{a^2+b}\sqrt{a+b^2}+\sqrt{ab+1}\geq 0$ . Then one of the variables is $\leq 0$ , one is $\geq 0$ . WLOG let $a\leq 0, b\geq 0$ . Then by $x=-1$ we get $(a\sqrt{b^2+a})^2=(-1-b\sqrt{a^2+b})^2$ , which rearranges to $a^3-b^3=1+2b\sqrt{a^2+b}$ . But since $a^3-b^3\leq 0$ and $\geq 0$ , this is impossible, so $\boxed{x=1}$ .

Side note, is a construction needed for this problem? The phrasing made it seem like you didn't, and I couldn't actually find one in contest (I think the exact phrasing is something like "Suppose real numbers $a, b$ satisfy...find, with prove, the value of...")

This post has been edited 1 time. Last edited by MortemEtInteritum, Mar 12, 2022, 4:35 AM

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Bluesoul

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Bluesoul

#3 h Mar 12, 2022, 4:38 AM • 1 Y

Y by ImSh95

Nice I got the answer $1$ as well
But I kinda guessed.

I cannot prove rigorously but when $a^2+b=0$ or $a+b^2=0$ is the only way the equation holds

Assume that $b>0,a<0$ and then I got $b^3=\frac{\sqrt{5}-1}{2}$ , the desired value is $\sqrt{b^6+b^3}=1$

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awesomeming327.

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awesomeming327.

#4 h Mar 12, 2022, 4:48 AM • 1 Y

Y by ImSh95

I somehow provided a wrong construction for both $x=1$ and also somehow thought $x=-1$ worked because I found a wrong construction.

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CANBANKAN

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CANBANKAN

#5 h Mar 12, 2022, 6:07 AM • 5 Y

Y by teomihai, Bluesoul, ImSh95, son7, rama1728

Bluesoul wrote:

Nice I got the answer $1$ as well
But I kinda guessed.

I cannot prove rigorously but when $a^2+b=0$ or $a+b^2=0$ is the only way the equation holds

Assume that $b>0,a<0$ and then I got $b^3=\frac{\sqrt{5}-1}{2}$ , the desired value is $\sqrt{b^6+b^3}=1$

This seems false....... and wildly unnecessary

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parmenides51

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parmenides51

#6 h Mar 19, 2022, 4:05 AM • 1 Y

Y by ImSh95

Assume that real numbers $a$ and $b$ satisfy $ab+\sqrt{ab+1}+\sqrt{a^2+b}\sqrt{a+b^2}=0.$ Find, with proof, the value of $b\sqrt{a^2+b}+a\sqrt{b^2+a}.$

This post has been edited 1 time. Last edited by Amir Hossein, Feb 13, 2023, 5:54 AM
Reason: Deleted unnecessary note.

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sqing

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sqing

#7 h Mar 19, 2022, 12:43 PM • 1 Y

Y by ImSh95

Let $a,b$ be real numbers satisfy $ab+\sqrt{ab+1}+\sqrt{(a^2+b)(a+b^2)}=0.$ Prove that $b\sqrt{a^2+b}+a\sqrt{b^2+a}=1.$
Good question.
2022CJMO

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sqing

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sqing

#9 h Mar 20, 2022, 2:21 AM • 1 Y

Y by ImSh95

sqing wrote:

Let $a,b$ be real numbers satisfy $ab+\sqrt{ab+1}+\sqrt{(a^2+b)(a+b^2)}=0.$ Prove that $b\sqrt{a^2+b}+a\sqrt{b^2+a}=1.$

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JingheZhang

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JingheZhang

#10 h Mar 20, 2022, 2:49 AM • 1 Y

Y by ImSh95

sqing wrote:

Let $a,b$ be real numbers satisfy $ab+\sqrt{ab+1}+\sqrt{(a^2+b)(a+b^2)}=0.$ Prove that $b\sqrt{a^2+b}+a\sqrt{b^2+a}=1.$

Where did you find this Chinese solution? Or you wrote this solution?

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sqing

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sqing

#11 h Mar 21, 2022, 1:25 AM • 1 Y

Y by ImSh95

Let $a,b\geq 0$ and $ab+\sqrt{3ab+1}+\sqrt{(a+b^2)(b+a^2)}=5.$ Prove that $2\sqrt 2\leq a\sqrt{a+b^2}+b\sqrt{b+a^2}\leq 4$

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LLL2019

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LLL2019

#12 h Mar 21, 2022, 4:47 AM • 1 Y

Y by ImSh95

My solution, I wrote up in last 5 minutes :blush:

so there were a lot of skipping. I wouldn't be surprised if this got a 0 or 1, though i may be slightly disappointed.

P.S. : Is there a collection made for this year's CMO yet?

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sqing

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sqing

#13 h Apr 8, 2022, 12:27 AM • 2 Y

Y by ImSh95, Dhruv777

Let $a,b\geq 0$ and $ab+\sqrt{3ab+1}+\sqrt{(a^2+b)(b^2+a)}=5.$ Prove that $2\sqrt 2\leq a\sqrt{a+b^2}+b\sqrt{b+a^2}\leq 4$

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sqing

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sqing

#14 h Apr 8, 2022, 3:40 AM • 1 Y

Y by ImSh95

sqing wrote:

Let $a,b\geq 0$ and $ab+\sqrt{3ab+1}+\sqrt{(a^2+b)(b^2+a)}=5.$ Prove that $2\sqrt 2\leq a\sqrt{a+b^2}+b\sqrt{b+a^2}\leq 4$

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mathaddiction

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mathaddiction

#17 h Apr 29, 2022, 8:11 AM • 4 Y

Y by ImSh95, mijail, TinSn, orangesyrup

Notice that
$\begin{align*} (b\sqrt{a^2+b}+a\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\sqrt{ab+1})\\ &=a^3+b^3-2ab\sqrt{ab+1}\\ &=a^3+b^3-(ab+\sqrt{ab+1})^2+a^2b^2+ab+1\\ &=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\\ &=1 \end{align*}$ Now obviously $ab<0$ , suppose $a>0>b$ , then $a^3>b^3$ which implies
$b^2(a^2+b)<a^2(b^2+a)$ Hence $|b\sqrt{a^2+b}|<|a\sqrt{b^2+a}|$ and so
$b\sqrt{a^2+b}+a\sqrt{b^2+a}>0$ which implies that the value is $1$ .

This post has been edited 2 times. Last edited by mathaddiction, Apr 30, 2022, 11:52 AM

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MortemEtInteritum

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MortemEtInteritum

#18 h Apr 29, 2022, 11:28 PM • 1 Y

Y by ImSh95

mathaddiction wrote:

Notice that
$\begin{align*} (b\sqrt{a^2+b}+a\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\sqrt{ab+1})\\ &=a^3+b^3-2ab\sqrt{ab+1}\\ &=a^3+b^3-(ab+\sqrt{ab+1})^2+a^2b^2+ab+1\\ &=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\\ &=1 \end{align*}$

A large part of the problem is actually proving whether the value is $1$ or $-1$

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mathaddiction

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mathaddiction

#19 h Apr 30, 2022, 11:53 AM • 1 Y

Y by ImSh95

MortemEtInteritum wrote:

mathaddiction wrote:

Notice that
$\begin{align*} (b\sqrt{a^2+b}+a\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\sqrt{ab+1})\\ &=a^3+b^3-2ab\sqrt{ab+1}\\ &=a^3+b^3-(ab+\sqrt{ab+1})^2+a^2b^2+ab+1\\ &=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\\ &=1 \end{align*}$

A large part of the problem is actually proving whether the value is $1$ or $-1$

Indeed! Thanks very much for pointing out, just edited.

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samrocksnature

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samrocksnature

#21 h May 3, 2022, 10:28 PM • 3 Y

Y by Hedra, chess12500, Mango247

Does this work???

Rewrite the given condition as $ab+\sqrt{a^2+b} \cdot \sqrt{b^2+a} = -\sqrt{ab+1}$ $a^3 + a b + 2 a^2 b^2 + b^3 + 2ab\sqrt{a^2 + b}\sqrt{a + b^2}=ab+1$ $a^3+ 2 a^2 b^2 + b^3 + 2ab\sqrt{a^2 + b}\sqrt{a + b^2}=1$ $(a\sqrt{b^2+a}+b\sqrt{a^2+b})^2=1$ $a\sqrt{b^2+a}+b\sqrt{a^2+b}= \pm 1$
We prove that it is equal to $1,$ or equivalently that it is positive. Since $a,b$ have opposite signs, assume $b$ is negative and let $b=-c$ for some positive real $c.$ $a\sqrt{b^2+a}+b\sqrt{a^2+b}=a\sqrt{c^2+a}-c\sqrt{a^2-c}.$ Note that we must have $a^2 \geq c$ or $a \geq \sqrt{c}.$ $a\sqrt{c^2+a}-c\sqrt{a^2-c} \geq \sqrt{c^3+c\sqrt{c}}\geq 0.$
Remarks

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Dhruv777

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Dhruv777

#22 h Sep 19, 2022, 1:19 AM

Y by

Any other approach to solve this problem?
:')

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lifeismathematics

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lifeismathematics

#23 h Nov 19, 2022, 1:42 PM

Y by

cute for algebra

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brainfertilzer

1831 posts

brainfertilzer

#24 h Feb 26, 2023, 6:29 PM

Y by

From the first equation,
$a^2b^2 + ab + 1 + 2ab\sqrt{ab + 1} = (a^2 + b)(b^2 + a) = a^2b^2 + b^3 + a^3 + ab,$ $1 + 2ab\sqrt{ab + 1} = a^3 +b^3.$ Let $k$ be the desired quantity. Then,
$k^2 = 2a^2b^2 + a^3 + b^3 + 2ab\sqrt{a^2 + b}\sqrt{b^2 + a},$ $k^2 = 2a^2b^2 + 2ab\sqrt{ab + 1} + 1 + 2ab(-ab - \sqrt{ab + 1}) = 1.$ Hence, $k = \pm 1$ . We claim $k = -1$ is impossible, so assume for contradiction $k = -1$ . Note that $ab \le 0$ , so $a\le 0\le b$ or $b\le 0\le a$ ; WLOG let it be the second one. We have $b\sqrt{a^2 + b} + a\sqrt{b^2 + a} = -1$ , and subtracting $a\sqrt{b^2 + a}$ and squaring gives
$b^2(a^2 + b) = 1 + a^2(b^2 + a) + 2a\sqrt{b^2 + a}$ $\implies b^3 = 1 + a^3 + 2a\sqrt{b^2 + a}\implies b^3 - a^3 = 1 + 2a\sqrt{b^2 +a}.$ But this is impossible because $b^3 - a^3\le 0$ and $1 + 2a\sqrt{b^2 +a}\ge 1$ . Hence, $k = +1$ .

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lamphead

503 posts

lamphead

#26 h Mar 4, 2024, 3:05 PM

Y by

samrocksnature wrote:

We prove that it is equal to $1,$ or equivalently that it is positive. Since $a,b$ have opposite signs, assume $b$ is negative and let $b=-c$ for some positive real $c.$ $a\sqrt{b^2+a}+b\sqrt{a^2+b}=a\sqrt{c^2+a}-c\sqrt{a^2-c}.$ Note that we must have $a^2 \geq c$ or $a \geq \sqrt{c}.$ $a\sqrt{c^2+a}-c\sqrt{a^2-c} \geq \sqrt{c^3+c\sqrt{c}}\geq 0.$

I may be misunderstanding some part, but the first part of the last inequality isn't necessarily true, right?. Consider a=3 and c=4, where $3\sqrt{19}-4\sqrt{5} \approx 4.13 < \sqrt{72} \approx 8.49$ .

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NicoN9

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NicoN9

#27 h Apr 12, 2025, 6:38 AM

Y by

I'm not sure if this is correct (especially the plus-minus thing, I'm bad at this type of algebra

) , but I think no one write in this approach so here is mine. (I was actually surprised that the answer is constant!)

Let $a\sqrt{a+b^2}+b\sqrt{a^2+b}=X$ . Consider two equations: $\begin{align*} \bullet (\sqrt{a^2+b}+a)(\sqrt{a+b^2}+b)&=\sqrt{a^2+b}\cdot \sqrt{a+b^2}+ab+X=X-\sqrt{ab+1} \\ \bullet (\sqrt{a^2+b}-a)(\sqrt{a+b^2}-b)&=\sqrt{a^2+b}\cdot \sqrt{a+b^2}+ab-X=-X-\sqrt{ab+1} \end{align*}$
Multiply these two: $\begin{align*} ab &=(\sqrt{ab+1}+X)(\sqrt{ab+1}-X) \\ &= ab+1-X^2 \end{align*}$ Therefore, $X=1$ or $X=-1$ .

From now, let's prove that $X\ge 0$ . WLOG $b$ is negative and let $c=-b$ . Then $a, c$ are positive and $\begin{align*} X &\ge 0 \\ \Longleftrightarrow a\sqrt{a^2+c} &\ge c\sqrt{a^2-c} \\ \Longleftrightarrow a^3 &\ge -c^3 \end{align*}$ which is obviously true.

This post has been edited 3 times. Last edited by NicoN9, Apr 12, 2025, 7:33 AM
Reason: fakesolve resolved

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