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Source: Izho 2025 P1

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116 posts

Iveela

#1 h Jan 14, 2025, 11:21 AM

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Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$ . Prove that $(a-b)^2 + a + b \geq 2$

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1198 posts

#2 h Jan 14, 2025, 11:39 AM • 4 Y

Y by VicKmath7, Rustem-E303, CahitArf, NicoN9

Trivial by $s = a + b$ and $p = ab$ . We have $s^3 - 3sp = p + 1$ , so $p = \frac{s^3-1}{3s + 1}$ (we cannot have $s = -\frac{1}{3}$ , as then $s^3 - 1 \neq 0$ ). From $p \leq \frac{s^2}{4}$ (equivalent to $(a-b)^2 \geq 0$ ), we get $s^3 - s^2 - 4 \leq 0$ , i.e. $(s-2)(s^2 + s + 2) \leq 0$ , so $s \leq 2$ . We wish to prove $s^2 - 4p + s \geq 2$ , i.e. $p \leq \frac{s^2 + s - 2}{4}$ . Equivalently, $\frac{s^3-1}{3s+1} \leq \frac{s^2+s-2}{4}$ , which rearranged to $s^3 - 4s^2 + 5s - 2 \leq 0$ , factoring as $(s-2)(s+1)^2 \leq 0$ , true by the abovementioned $s\leq 2$ .

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arqady

#3 h Jan 14, 2025, 11:40 AM

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Iveela wrote:

Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$ . Prove that $(a-b)^2 + a + b \geq 2$

See here

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1198 posts

#4 h Jan 14, 2025, 11:42 AM

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arqady wrote:

Iveela wrote:

Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$ . Prove that $(a-b)^2 + a + b \geq 2$

See here

Yeah, I saw that the problem was somehow leaked during the competition, better to ignore/delete that post.

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#5 h Jan 14, 2025, 12:06 PM

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"I’m not leaked. I just saw Izho’s problems (my friend sent them to me), and I thought the time was over, but a guy told me it wasn’t. then I tried to delete my two posts." ( my fault)

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1198 posts

#6 h Jan 14, 2025, 12:11 PM

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Titibuuu wrote:

"I’m not leaked. I just saw Izho’s problems (my friend sent them to me), and I thought the time was over, but a guy told me it wasn’t. then I tried to delete my two posts." ( my fault)

Ok, no worries, accidents happen. Can you please post Problem 2 again? (it is the only one from today currently not posted)

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#7 h Jan 14, 2025, 12:25 PM

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Assassino9931 wrote:

Titibuuu wrote:

"I’m not leaked. I just saw Izho’s problems (my friend sent them to me), and I thought the time was over, but a guy told me it wasn’t. then I tried to delete my two posts." ( my fault)

Ok, no worries, accidents happen. Can you please post Problem 2 again? (it is the only one from today currently not posted)

Thank you for understanding me, and I'm sorry again. I will
post now

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42 posts

#8 h Jan 14, 2025, 12:40 PM

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I apologize

if the Olympiad in which Izho participated gave the impression that it was unfair, and I hope that nothing like this has happened

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#9 h Jan 14, 2025, 12:46 PM • 3 Y

Y by pavel kozlov, HotSinglesInYourArea, Calamarul

It seems there are $2$ posts with this problem. Here is the solution I've posted on the other one as well:

By AM-GM we have $ab+2=a^3+b^3+1\geq 3ab\implies ab\leq 1$

$a^3+b^3=ab+1\implies (a+b)(a^2-ab+b^2)=ab+1\implies a^2-ab+b^2=\dfrac{ab+1}{a+b}$

$(a-b)^2+a+b\geq 2\iff a^2+b^2-ab+a+b-ab\geq 2\iff \dfrac{ab+1}{a+b}+a+b-ab\geq 2\iff$
$\iff (a+b)^2-ab(a+b)+ab+1\geq 2(a+b)\iff (a+b-1)^2+ab(1-a-b)\geq 0\iff$
$\iff -(a-1)(b-1)(a+b-1)\geq 0\iff (a-1)(b-1)(a+b-1)\leq 0$

If $a\geq 1\implies b\leq 1$ (since $ab\leq 1$ ) and $a+b>1$ , so the inequality holds.

Similarly, if $b\geq 1$ the inequality holds.

Claim: The equation $a^3+b^3=ab+1$ doesn't have any solutions in $(0,1)^2$ .

Proof:

Suppose there exists $(a,b)\in(0,1)^2$ .

$a, b<1\implies a^3<a$ and $b^3<b$ , so $1=a^3+b^3-ab<a+b-ab\implies ab-a-b+1<0\implies (a-1)(b-1)<0$ , which is clearly false and we're done.

This post has been edited 1 time. Last edited by Double07, Jan 14, 2025, 12:52 PM

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maths_enthusiast_0001

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maths_enthusiast_0001

#10 h Jan 14, 2025, 12:57 PM

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Here is my solution from a thread earlier. Apparently #2 is literally same as this.
Let $a+b=u$ and $ab=v$ ( $u,v \in \mathbb{R^{+}}$ since $a,b \in \mathbb{R^{+}}$ )
By AM-GM Inequality,
$a^3+b^3+1 \geq 3ab$ $\implies ab+2 \geq 3ab \implies \boxed{v \leq 1}$ By Power-Mean Inequality,
${\left(\frac{a^3+b^3+1}{3}\right)}^{1/3} \geq \left(\frac{a+b+1}{3}\right)$ $\implies {\left(\frac{ab+2}{3}\right)}^{1/3} \geq \left(\frac{a+b+1}{3}\right) \implies \frac{u+1}{3} \leq {\left(\frac{v+2}{3}\right)}^{1/3} \leq 1(\because v \leq 1)$ $\implies \boxed{u \leq 2}$ Now, $a^3+b^3=ab+1 \implies (a+b)^3-3ab(a+b)=(ab+1) \implies u^3-3uv=v+1 \implies \boxed{v=\frac{u^3-1}{3u+1}}$ . We wish to prove that
$(a-b)^2+a+b \geq 2$ $\iff (a+b)^2+(a+b)-4ab \geq 2 \iff u^2+u-4v \geq 2$ $\iff (u^2+u)-4\left(\frac{u^3-1}{3u+1}\right) \geq 2$ $\iff (u^2+u)(3u+1)-4(u^3-1) \geq 2(3u+1)$ $\iff -u^3+4u^2-5u+2 \geq 0 \iff \boxed{(u-1)^{2}(2-u) \geq 0}$ which is indeed true. $\blacksquare$ ( $\mathcal{QED}$ )

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867 posts

#11 h Jan 14, 2025, 2:47 PM • 16 Y

Y by Assassino9931, NO_SQUARES, MS_asdfgzxcvb, VicKmath7, turann, Rustem-E303, ehuseyinyigit, sami1618, Osim_09, Mathlover_1, yobu, alexanderhamilton124, pavel kozlov, ytChen, Stuffybear, NicoN9

Here is the shortest proof:

$a^2+b+b^2+a\ge 2\sqrt{(a^2+b)(b^2+a)}=2(ab+1)$

This post has been edited 1 time. Last edited by rightways, Jan 14, 2025, 2:52 PM
Reason: Eng grammar typo

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sqing

#12 h Jan 14, 2025, 4:03 PM • 1 Y

Y by rightways

Let $a,b\geq 0$ and $a^3 + b^3 - ab\geq 1.$ Prove that
$(a - b)^2 + a + b \geq 2$

This post has been edited 1 time. Last edited by sqing, Jan 14, 2025, 4:21 PM

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41170 posts

sqing

#13 h Jan 14, 2025, 4:22 PM • 1 Y

Y by Wiley_H

rightways wrote:

Here is the shortest proof:

$a^2+b+b^2+a\ge 2\sqrt{(a^2+b)(b^2+a)}=2(ab+1)$

Very nice.

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#14 h Jan 20, 2025, 2:46 PM

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Can we use vieta formulas in this problem ? I try to find alternative solution

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#15 h Jan 21, 2025, 10:06 AM

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let $f(a,b)=a^2-2ab+a+b^2+b-2$ and $g(a,b)=a^3-ab+b^3-1$ . now we can use the method of Lagrange multipliers; in the minimum point of $f$ we should have:
$\nabla f = \lambda \nabla g \Longleftrightarrow 2a-2b+1 = \lambda (3a^2 - b) \quad \& \quad 2b-2a+1 = \lambda (3b^2-a)$ .
if $3b^2 = a$ then $a = b + \frac{1}{2}$ so $3b^2 - b - \frac{1}{2} = 0$ but this equation doesn't have any real root ,contradiction!. so we can write equations in this form:
$\frac{2a-2b+1}{3a^2-b}=\frac{2b-2a+1}{3b^2-a} \Longleftrightarrow (a-b)(6a^2+6b^2-5a-5b-1)=0$
in the case $a=b=x$ we have $2x^3-x^2-1=(x-1)(2x^2 + x + 1)=0$ the second term doesn't have positive root so we get $a=b=1$ and it's easy to see they work!
so we can assume that $h(a,b) = 6a^2+6b^2-5a-5b-1 = 0$ . now we found out in minimum point of $f$ with knowing $g=0$ then $h=0$ . we can prove minimum of $f$ on $h$ is bigger than 0 and it proves the problem!!. for this we use the method of Lagrange multipliers again!!.
$\nabla f = \lambda \nabla h \Longleftrightarrow 2a-2b+1 = \lambda (12a-5) \quad \& \quad 2b-2a+1 = \lambda (12b-5)$
so $4(a-b) = 12\lambda (a-b)$ then with knowing $a \neq b$ we have $\lambda = \frac{1}{3}$ and we get $2b-2a+1 = 4b - \frac{5}{3} \Longleftrightarrow a+b = \frac{4}{3}$ applying on $h=0$ we get:
$h(a,b,c)= 6(a+b)^2 -12ab - 5(a+b) -1 = 3 - 12ab = 0 \Longleftrightarrow ab = \frac{1}{4}$ . so we have to prove $f$ is positive on the point $(a,b)$ and it's:
$f(a,b) = (a+b)^2 - 4ab +(a+b) - 2 = \frac{1}{9} > 0$ . and the problem is done!

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#16 h Jan 22, 2025, 7:26 AM

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Really easy

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41170 posts

sqing

#17 h Jan 22, 2025, 11:17 AM

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Let $a, b$ be reals such that $a^2+b^2+ a^5+ b^5 = ab + 3$ . Prove that $(a-b)^2 + a + b \geq 2$ Let $a, b$ be reals such that $a^3 + b^3+ a^5+ b^5 = ab +3$ . Prove that $(a-b)^2 + a + b \geq 2$ Let $a, b$ be reals such that $a^2+b^2+a^3 + b^3+ a^5+ b^5 = ab + 5$ . Prove that $(a-b)^2 + a + b \geq 2$ h

This post has been edited 1 time. Last edited by sqing, Jan 22, 2025, 11:38 AM

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NicoN9

#18 h Mar 20, 2025, 2:32 AM

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WLOG $a\ge b$ . Furthermore, if $a=b$ , then we easily get $a=1$ . Thus, we may assume $a>b$ .

Claim. There are no solution when $1\ge a >b$ .
$proof$ . Assume $a\le 1$ . now, since $a^3+b^3<a^2+b^2$ , we have $a^2+b^2-ab\ge 1\Longleftrightarrow a^3+b^3\ge a+b$ as desired. Here, we use the fact that $a^3+b^3=(a+b)(a^2-ab+b^2)$ .

Note that $a^3+b^3=\frac{ab+1}{a+b}$ . Now, the problem inequality is equivalent to $\begin{align*} (a^2+b^2-ab)+(a+b-ab)\geq 2\\ \Longleftrightarrow \frac{ab+1}{a+b}-1-(ab-a-b+1)\geq 0\\ \Longleftrightarrow (a-1)(b-1)(1-a-b)\geq 0. \end{align*}$
We casework here wrt the value of $a, b$ .
case1: $a\ge b\ge 1$
trivial.
case2: $a\ge 1\ge b$
inequality above makes trivial.
case 3: $1\ge a >b$
By the claim, this is impossible.
as these cover all the cases, we are done.

This post has been edited 1 time. Last edited by NicoN9, Mar 20, 2025, 2:34 AM

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sqing

#19 h Mar 20, 2025, 3:23 AM

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Let $a,b\geq 0$ and $a^3 + b^3 +a^2b^2- ab \geq2.$ Prove that
$(a - b)^2 + a + b \geq 2$

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