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So the question I've been trying to solve is :

for $(x, y)$ in $R²$ $x^2 + y^2 + xy = x + y \Rightarrow 0 \le x + y \le \frac{4}{3}$

I've tried two things so far :

  • I have put $t = (x+y)$ so the first equation becomes $t² -t -xy = 0$ this means that $1 + 4xy \ge 0$ , I've tried some things with this but nothing.
  • I've tried proving that for $(x,y)$ in $R²$ $x + y > \frac{4}{3} $ OR $ x + y < 0 => x² + y² + xy$ is different that $x + y$, since ($P=>Q$) is the same thing as (not Q $=>$ not P)
mathlove
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enzo
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1 Answers1

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Let $x+y=a,xy=b$. Then, we have $$a^2-b=a,$$ i.e. $$b=a^2-a.$$

By the way, since $x,y$ are the real roots of $t^2-at+b=0$, we have $$(-a)^2-4\cdot 1\cdot b\ge 0,$$ i.e. $$a^2-4(a^2-a)\ge 0,$$ i.e. $$0\le a\le\frac 43.$$

mathlove
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  • can you please add more steps between the roots of t²−at+b=0 and the next steps ? – enzo Oct 21 '15 at 17:24
  • @maetsoh: The roots are real, so the discriminant has to be non-negative. – mathlove Oct 21 '15 at 17:25
  • @maetsoh: $x,y$ are the real roots of $(t-x)(t-y)=0,$ i.e. $t^2-(x+y)t+xy=0,$ i.e. $t^2-at+b=0$. – mathlove Oct 21 '15 at 17:41
  • what is t ? is t = x + y ? – enzo Oct 21 '15 at 18:06
  • @maetsoh: No. I just used $t$ to express an equation which has $x,y$ as the real roots. There is another way to get $a^2-4b\ge 0$ : you can have $(x-y)^2=(x+y)^2-4xy=a^2-4b$. Since $(x-y)^2$ is non-negative, we have $a^2-4b\ge 0$. – mathlove Oct 21 '15 at 18:11
  • I'll try what you suggested, because I've already tried the thing with the discriminant for x² + y² + xy = x + y, I've put t = x + y the equation became t² − t − xy = 0, and it gave me 1+4xy ≥ 0 – enzo Oct 21 '15 at 18:21
  • @maetsoh: OK. Then, do you have anything unclear? Note that $t$ is not important, you can write $t^2-at+b=0$ as $p^2-ap+b=0,q^2-aq+b=0$ or $s^2-as+b=0$. In any case, the equation has $x,y$ as the real roots, so the discriminant $(-a)^2-4b$ has to be non-negative. – mathlove Oct 21 '15 at 18:25
  • why do we suppose that x,y are roots of t² − at + b = 0 ? – enzo Oct 21 '15 at 18:30
  • @maetsoh: You agree that $x,y$ are the real roots of an equation $(t-x)(t-y)=0$, don't you? Then, expanding the left hand side gives $(t-x)(t-y)=t^2-ty-tx+xy=t^2-(x+y)t+xy=t^2-at+b$, right? So, $t^2-at+b=0$ has $x,y$ as the real roots. – mathlove Oct 21 '15 at 18:32
  • yes I totally agree with what you've written but what's the connection between this and the equation in my question ? $x^2+y^2+xy=x+y$ . – enzo Oct 21 '15 at 18:37
  • @maetsoh: From the equation in your question, we have $b=a^2-a$, right? And since $x,y$ are the real roots of $t^2-at+b=0$, we have that the discriminant $(-a)^2-4b$ has to be non-negative, i.e. $(-a)^2-4b\ge 0$. Now since $b=a^2-a$, we have $(-a)^2-4(a^2-a)\ge 0$, now solve this. – mathlove Oct 21 '15 at 18:40
  • oh I feel dumb. Thanks a lot – enzo Oct 21 '15 at 18:45