<s>[INST] Finding specific interval The question is to find $[a,b]$ such that $$\forall x,y : x\in [a,\frac {a+b}{2}],y\in[\frac{a+b}{2},b]\Rightarrow x+y,xy \in [a,b]$$
I tried like below
$$a \leq x \leq\frac {a+b}{2}\\
\frac {a+b}{2} \leq y \leq b\\
\to \frac {3a+b}{2} \leq x+y \leq \frac {a+3b}{2}\\\to 
a\leq \frac {3a+b}{2}\leq \frac {a+3b}{2}\leq b$$ this all get $a\leq b $
and $$a.\frac {a+b}{2} \leq xy \leq b.\frac {a+b}{2}\\
a\leq a.\frac {a+b}{2} \leq b.\frac {a+b}{2}\leq b$$ for a>o will be $$2a\leq a^2+ab \to 2\leq a+b\\
b.\frac {a+b}{2} \leq b \to a+b\leq 2$$
this mean $a+b=2$
first :Am I on right track?
second: I got stuck on this,because It revert to trivially $a<b$
Please guide me to find the solution? or give an Idea to take over
thanks in advance
 [/INST] $$a \le \frac{3a+b}2 \iff 0 \le a+b$$
$$\frac{a+3b}{2}\le b \iff a + b \le 0$$
Hence we have $a+b=0$.
$$a=-b$$
$|x| \le b$, $|y| \le b$, we want $|xy| \le b$
$$b^2 \le b$$
$$b^2-b \le 0$$
$$b(b-1) \le 0$$
Hence,
$$0 \le b \le 1, a = -b$$
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