Let $xy=a$. We have $(x+y)^2\geq0\Rightarrow 2\geq x^2+y^2\geq -2xy=-2a\Rightarrow a\geq-1\Rightarrow a+1\geq0.$ $$\Rightarrow xy+3=a+1+2\geq2\sqrt{2(a+1)}=2\sqrt{2a+2}=2\sqrt{2xy+2}\geq2\sqrt{2xy+x^2+y^2}=2|x+y|\geq2(x+y)$$

With $s=x+y$, $p=xy$ we have $s^2 - 2p \leq 2$, i.e. $p \geq \frac{s^2}{2} - 1$ and wish to show $p+3 \geq 2s$, i.e. $p \geq 2s-3$. Hence it is enough to prove $\frac{s^2}{2} - 1 \geq 2s-3$, but this is equivalent to $(s-2)^2\geq 0$.

Let $ x, y $ be real numbers such that $x^2 + y^2 \le 2$. Prove that $$-5 \le 2x + 2y-xy\le 3$$$$-4\le x + y-2xy\le \frac{9}{4} $$$$-3 \le x + y-xy\le\frac{3}{2} $$

EmirhanYagcioglu wrote: Let $xy=a$. We have $(x+y)^2\geq0\Rightarrow 2\geq x^2+y^2\geq -2xy=-2a\Rightarrow a\geq-1\Rightarrow a+1\geq0.$ $$\Rightarrow xy+3=a+1+2\geq2\sqrt{2(a+1)}=2\sqrt{2a+2}=2\sqrt{2xy+2}\geq2\sqrt{2xy+x^2+y^2}=2|x+y|\geq2(x+y)$$ good one

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Assassino9931 wrote: With $s=x+y$, $p=xy$ we have $s^2 - 2p \leq 2$, i.e. $p \geq \frac{s^2}{2} - 1$ and wish to show $p+3 \geq 2s$, i.e. $p \geq 2s-3$. Hence it is enough to prove $\frac{s^2}{2} - 1 \geq 2s-3$, but this is equivalent to $(s-2)^2\geq 0$. beautiful

Let $ x, y $ be real numbers such that $x^2 +x+y+ y^2 \le 2$. Prove that $$- \frac{51+3\sqrt 5}{2} \le x + y-xy\le \frac{9}{8} $$$$- 2(2+\sqrt 5) \le x + y-2xy\le 2$$$$- \frac{5\sqrt 5+7}{2} \le 2x +2 y-xy\le\frac{5\sqrt 5-7}{2} $$