wrong solution

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 $(x,y) = \pm{(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})}$

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 $x=y=\pm \textstyle\frac{1}{\sqrt{3}}$.

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.

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:

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 $x^2=\frac{1}{3}\implies x=y=\pm \frac{\sqrt{3}}{3}$

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

$X=Y=\frac1{\sqrt3}$ 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 $X=Y=\frac1{\sqrt3}$.

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*}

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.