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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
3 var inequality
SunnyEvan   13
N 39 minutes ago by Nguyenhuyen_AG
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
13 replies
+1 w
SunnyEvan
May 17, 2025
Nguyenhuyen_AG
39 minutes ago
trigonometric inequality
MATH1945   13
N an hour ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
13 replies
MATH1945
May 26, 2016
sqing
an hour ago
Iran TST Starter
M11100111001Y1R   2
N an hour ago by sami1618
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
2 replies
M11100111001Y1R
Tuesday at 7:36 AM
sami1618
an hour ago
Twin Prime Diophantine
awesomeming327.   23
N 2 hours ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
1 viewing
awesomeming327.
Mar 7, 2025
HDavisWashu
2 hours ago
No more topics!
symmetric algebra
Bluesoul   21
N Apr 12, 2025 by NicoN9
Source: 2022 CMO
If $ab+\sqrt{ab+1}+\sqrt{a^2+b}\sqrt{a+b^2}=0$, find the value of $b\sqrt{a^2+b}+a\sqrt{b^2+a}$
21 replies
Bluesoul
Mar 12, 2022
NicoN9
Apr 12, 2025
symmetric algebra
G H J
G H BBookmark kLocked kLocked NReply
Source: 2022 CMO
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Bluesoul
899 posts
#1 • 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
1332 posts
#2 • 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
899 posts
#3 • 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.
1737 posts
#4 • 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
1301 posts
#5 • 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
30653 posts
#6 • 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
42488 posts
#7 • 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
Attachments:
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sqing
42488 posts
#9 • 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
179 posts
#10 • 1 Y
Y by ImSh95
sqing wrote:
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
42488 posts
#11 • 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
834 posts
#12 • 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?
Attachments:
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sqing
42488 posts
#13 • 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
42488 posts
#14 • 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
308 posts
#17 • 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
1332 posts
#18 • 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
308 posts
#19 • 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. :D
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samrocksnature
8791 posts
#21 • 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
8 posts
#22
Y by
Any other approach to solve this problem?
:')
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lifeismathematics
1188 posts
#23
Y by
cute for algebra
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brainfertilzer
1831 posts
#24
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, $k = +1$.

Comment
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lamphead
503 posts
#26
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
158 posts
#27
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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