Proving that if $x^2+y^2+xy=x+y$, then $0\le x+y\le \frac 43$ for $x,y\in\mathbb R$.

Question

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)

Answer 1

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. $$0\le a\le\frac 43.$$