Find the optimal constant C in an inequality

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Find the optimal constant C in an inequality

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Source: Austrian MO 2024, Final Round P1

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#1 h Jun 1, 2024, 12:24 PM

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Determine the smallest real constant $C$ such that the inequality
$(X+Y)^2(X^2+Y^2+C)+(1-XY)^2 \ge 0$ holds for all real numbers $X$ and $Y$ . For which values of $X$ and $Y$ does equality hold for this smallest constant $C$ ?

(Walther Janous)

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#2 h Jun 1, 2024, 1:14 PM

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wrong solution

This post has been edited 1 time. Last edited by sami1618, Jun 1, 2024, 1:53 PM

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#3 h Jun 1, 2024, 1:30 PM

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i claim $C= -1$ is the solution now let $C = -1-r$ where $r>0$ we intend to try to find a smaller C
$(x^2+y^2)(x+y)^2+(-1-r)(x+y)^2+(1-xy)^2 \geq 0$
$(x^2+y^2)(x+y)^2-(x+y)^2-r(x+y)^2+(1-xy)^2\geq 0$
$(x^2+y^2-1)(x+y)^2+(1-xy)^2 -r(x+y)^2 \geq 0$

Claim
$(x^2+y^2-1)(x+y)^2+(1-xy)^2 \geq 0$ for all $x,y \in R$

Proof:

$(x^2+xy+y^2)^2+(x-y)^2-2xy+1\geq x^2y^2+(x-y)^2-2xy+1 = (xy-1)^2+(x-y)^2 \geq 0$
this is true since
$(x^2+y^2+xy-xy)(x^2+y^2+2xy) \geq 0$
$(x+y)^2(x^2+y^2) \geq 0$

so if $r >0$ and $(x+y) \ne 0$ thus it will become negative forcing the $\lim_{r\rightarrow 0} r$
GG at above, It is quite hard to find the SOS, but ur equality case is wrong, I find the $(x,y) = \pm{(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})}$

This post has been edited 1 time. Last edited by StarLex1, Jun 1, 2024, 1:34 PM
Reason: edit

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#4 h Jun 5, 2024, 12:00 PM

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StarLex1 wrote:

Claim
$(x^2+y^2-1)(x+y)^2+(1-xy)^2 \geq 0$ for all $x,y \in R$
Proof:
$(x^2+xy+y^2)^2+(x-y)^2-2xy+1\geq x^2y^2+(x-y)^2-2xy+1 = (xy-1)^2+(x-y)^2 \geq 0$
...

I don't understand this solution, it seems that $(x^2+y^2-1)(x+y)^2+(1-xy)^2 \not= (x^2+xy+y^2)^2+(x-y)^2-2xy+1$ .

As for the proof in the case $C=-1$ : Write $s=x+y$ , $p=xy$ , then we have
$(x^2+y^2-1)(x+y)^2+(1-xy)^2 = (s^2-2p-1)s^2+(1-p)^2=p^2-2(s^2+1)p+(s^4-s^2+1) = (p - (s^2+1))^2 - 3s^2.$ Now since $p \le \textstyle \frac{s^2}{4}$ , we obtain
$(p - (s^2+1))^2 - 3s^2 \ge \left( \textstyle\frac{3s^2}{4}+1 \right)^2 - 3s^2 =\left( \textstyle\frac{3s^2}{4}-1 \right)^2 \ge 0.$ Equality holds iff $s^2=4p = \textstyle\frac{4}{3}$ , which means that $x=y=\pm \textstyle\frac{1}{\sqrt{3}}$ .

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NumberzAndStuff

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#5 h Aug 30, 2024, 10:49 AM

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We claim C = -1 is the solution. Observe:
$(X+Y)^2(X^2+Y^2-1) + (1-XY)^2 = (X-Y)^2 + (X^2+Y^2+XY-1)^2 \geq 0$
Since there are equality cases, we are done.

This post has been edited 3 times. Last edited by NumberzAndStuff, Sep 21, 2024, 4:17 PM
Reason: latex

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sqing

#6 h Sep 25, 2024, 1:44 PM

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Tintarn wrote:

Determine the smallest real constant $C$ such that the inequality
$(X+Y)^2(X^2+Y^2+C)+(1-XY)^2 \ge 0$ holds for all real numbers $X$ and $Y$ . For which values of $X$ and $Y$ does equality hold for this smallest constant $C$ ?

(Walther Janous)

$(x+y)^2(x^2+y^2-1)+(1-xy)^2 \geq 0$

Attachments:

This post has been edited 1 time. Last edited by sqing, Sep 25, 2024, 1:45 PM

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#7 h Oct 1, 2024, 11:17 AM

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Let $P(x,y)=(x+y)^2(x^2+y^2+c)+(1-xy)^2\geq0$
$P(\sqrt{x},\sqrt{x})=9x^2+x(4c-2)+1\geq0 \forall x,y\in \mathbb{R}$ But that only works for $c\in [-1, 2]$
Therefore, $2 \geq c \geq -1$ , $c=-1$ holds for $x^2=\frac{1}{3}\implies x=y=\pm \frac{\sqrt{3}}{3}$

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#8 h Oct 1, 2024, 3:53 PM

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Progress so far: $c \ge -1$ .
Trying to prove $c = -1$ works using Sum of Squares, but it isn't working.

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DottedCaculator

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#9 h Oct 1, 2024, 4:55 PM

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$X=Y=\frac1{\sqrt3}$ implies $C\geq-1$ . This works because $(X+Y)^2(X^2+Y^2-1)+(1-XY)^2=(X^2+XY+Y^2)^2+1-(X^2+4XY+Y^2)\geq2(X^2+XY+Y^2)-(X^2+4XY+Y^2)=(X-Y)^2=0$ . Equality holds if and only if $X=Y=\frac1{\sqrt3}$ .

This post has been edited 1 time. Last edited by DottedCaculator, Oct 1, 2024, 4:56 PM

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#10 h Oct 3, 2024, 4:48 PM • 1 Y

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Consider the cotangent function.
Since $a^2+b^2+c^2\geq ab+ac+bc,$ then
$\begin{align*} X^2+Y^2+\dfrac{(1-XY)^2}{(X+Y)^2}\geq X\cdot \dfrac{1-XY}{X+Y}+Y\cdot \dfrac{1-XY}{X+Y}+X\cdot Y=1. \end{align*}$

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Victoria_Discalceata1

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#11 h Oct 4, 2024, 2:05 AM

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Tintarn wrote:

Determine the smallest real constant $C$ such that the inequality
$(X+Y)^2(X^2+Y^2+C)+(1-XY)^2 \ge 0$ holds for all real numbers $X$ and $Y$ . For which values of $X$ and $Y$ does equality hold for this smallest constant $C$ ?

(Walther Janous)

We have $0\le (X+Y)^2\left(X^2+Y^2+C\right)+(1-XY)^2=\frac{1}{4}\left(X^2+4XY+Y^2-2\right)^2+\frac{3}{4}\left(X^2-Y^2\right)^2+(C+1)(X+Y)^2$ hence if $C\ge -1$ then the inequality holds for all real $X,Y$ . On the other hand, take $C=-1-\varepsilon$ , where $\varepsilon >0$ , and $X=Y=\frac{1}{\sqrt{3}}$ to obtain $0\le -\frac{4\varepsilon}{3}$ , which is a contradiction.

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