It's $(xy+1)^2+(x+y-2)^2\geq 0$ which is true. Equality cases when :$xy=-1,x+y=2$ which gives$(1+\sqrt{2};1-\sqrt{2})$

GGPiku wrote: It's $(xy+2)^2+(x+y-2)^2\geq 0$ How did you get here?

cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. Very nice. There is a simple proof.

Not_a_Username wrote: GGPiku wrote: It's $(xy+2)^2+(x+y-2)^2\geq 0$ How did you get here? Had a little typo, sorry.

cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. Proof of Zhangyunhua:

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cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. Good practice of completing square of some quadratic function in $x$ for Japanese Kids at age of 13, 14, 15. $(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1)=(y^2+1)x^2+4(y-1)x+y^2-4y+5$ $=(y^2+1)\left\{x+\frac{4(y-1)}{y^2+1}\right\}^2+y^2-4y+5-\frac{4(y-1)^2}{y^2+1}$ $=(y^2+1)\left\{x+\frac{2(y-1)}{y^2+1}\right\}^2+\frac{y^4-4y^3+2y^2+4y+1}{y^2+1}$ $=(y^2+1)\left\{x+\frac{2(y-1)}{y^2+1}\right\}^2+\frac{(y^2-2y-1)^2}{y^2+1}\ge 0$ Equality holds if only if $x+\frac{2(y-1)}{y^2+1}=0$ and $y^2-2y-1=0$ $\Longleftrightarrow y=1\pm \sqrt{2}$ and $x=-\frac{2(y-1)}{2(y+1)}=\frac{2}{y+1}-1$ $\therefore (x, \ y)=(1-\sqrt{2},\ 1+\sqrt{2}),\ (1+\sqrt{2},\ 1-\sqrt{2}).$

Kunihiko_Chikaya wrote: cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. Good practice of completing square of some quadratic function in $x$ for Japanese Kids at age of 13, 14, 15. $(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1)=(y^2+1)x^2+4(y-1)x+y^2-4y+5$ $=(y^2+1)\left\{x+\frac{4(y-1)}{y^2+1}\right\}^2+y^2-4y+5-\frac{4(y-1)^2}{y^2+1}$ $=(y^2+1)\left\{x+\frac{2(y-1)}{y^2+1}\right\}^2+\frac{y^4-4y^3+2y^2+4y+1}{y^2+1}$ $=(y^2+1)\left\{x+\frac{2(y-1)}{y^2+1}\right\}^2+\frac{(y^2-2y-1)^2}{y^2+1}\ge 0$ Equality holds if only if $x+\frac{2(y-1)}{y^2+1}=0$ and $y^2-2y-1=0$ $\Longleftrightarrow y=1\pm \sqrt{2}$ and $x=-\frac{2(y-1)}{2(y+1)}=\frac{2}{y+1}-1$ $\therefore (x, \ y)=(1-\sqrt{2},\ 1+\sqrt{2}),\ (1+\sqrt{2},\ 1-\sqrt{2}).$ Wonderful.

cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. $(x^2+1)(y^2+1)=(x+yi)(x-yi)(y+i)(y-i)=(x+i)(y+i)\times (x-i)(y-i)$ $=\{(xy-1)+(x+y)i\}\{(xy-1)-(x+y)i\}$ $=(xy-1)^2+(x+y)^2$, thus we have $(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1)=(xy-1)^2+(x+y)^2+4\{xy-(x+y)+1\}$ $=(x+y-2)^2+(xy+1)^2\geq 0.$ Equality holds if only if $x+y=2$ and $xy=-1$, that is to say, $x,y$ are the roots of the quadratic equation in $t$, $t^2-2t-1=0\Longleftrightarrow t=1\pm \sqrt{2}$, yielding $(x,\ y)=(1-\sqrt{2},\ 1+\sqrt{2}),\ (1+\sqrt{2},\ 1-\sqrt{2}).$

Kunihiko_Chikaya wrote: cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. $(x^2+1)(y^2+1)=(x+yi)(x-yi)(y+i)(y-i)=(x+i)(y+i)\times (x-i)(y-i)$ $=\{(xy-1)+(x+y)i\}\{(xy-1)-(x+y)i\}$ $=(xy-1)^2+(x+y)^2$, thus we have $(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1)=(xy-1)^2+(x+y)^2+4\{xy-(x+y)+1\}$ $=(x+y-2)^2+(xy+1)^2\geq 0.$ Equality holds if only if $x+y=2$ and $xy=-1$, that is to say, $x,y$ are the roots of the quadratic equation in $t$, $t^2-2t-1=0\Longleftrightarrow t=1\pm \sqrt{2}$, yielding $(x,\ y)=(1-\sqrt{2},\ 1+\sqrt{2}),\ (1+\sqrt{2},\ 1-\sqrt{2}).$ $(x^2+1)(y^2+1)=(x+i)(y+i)\times (x-i)(y-i)$ $=\{(xy-1)+(x+y)i\}\{(xy-1)-(x+y)i\}=(xy-1)^2+(x+y)^2$ Very nice.

cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. The problem seemed to be Very Easy. $(x^2+1)(y^2+1)+4(x-1)(y-1)=x^2y^2+x^2+y^2+1+4\{xy-(x+y)+1\}$ $=x^2y^2+(x+y)^2-2xy+1+4xy-4(x+y)+4$ $=x^2y^2+2xy+1+(x+y)^2-4(x+y)+4$ $=(xy+1)^2+(x+y-2)^2\geq 0.$ Equality holds if only if $x+y=2$ and $xy=-1$, that is to say, $x,y$ are the roots of the quadratic equation in $t$, $t^2-2t-1=0\Longleftrightarrow t=1\pm \sqrt{2}$, yielding $(x,\ y)=(1-\sqrt{2},\ 1+\sqrt{2}),\ (1+\sqrt{2},\ 1-\sqrt{2}).$

cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. $(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1)=(y^2+1)x^2+4(y-1)x+y^2-4y+5.$ $D/4=\{2(y-1)\}^2-(y^2+1)(y^2-4y+5)$ $=-(y^4-4y^3+2y^2+4y+1)=-(y^2-2y-1)^2\leq 0$, which shows the proof. Equality holds if only if $y^2-2y-1=0$ and $x=-\frac{2(y-1)}{y^2+1}$ $\Longleftrightarrow y=1\pm \sqrt{2}$ and $x=-\frac{2(y-1)}{2(y+1)}=\frac{2}{y+1}-1$ $\therefore (x, \ y)=(1-\sqrt{2},\ 1+\sqrt{2}),\ (1+\sqrt{2},\ 1-\sqrt{2}).$

cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. https://cms.math.ca/Competitions/REP/2017/2017CMOQR_solutions_en.pdf

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$$xy=a,x+y=b$$$$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$$$(a+1)^2+(b-2)^2 \geq 0.$$Equality case $a=-1,b=2$

cjquines0 wrote: Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$Determine when equality holds. beautiful.

Easy ! $(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$ $x^2y^2+x^2+y^2+1+4xy-4y-4x+4 \geq 0 $ $x^2y^2+(x+y)^2+2xy-4x-4y+5 \geq 0$ $ (x^2y^2+1+2xy)+(x+y)^2-4(x+y)+4 = (xy+1)^2+((x+y)-2)^2 \geq 0 $

GGPiku wrote: It's $(xy+1)^2+(x+y-2)^2\geq 0$ which is true. Equality cases when :$xy=-1,x+y=2$ which gives$(1+\sqrt{2};1-\sqrt{2})$ see here