Devastator wrote: Given $a,b,c \in \mathbb{R}$ satisfying $a+b+c=a^2+b^2+c^2=1$, show that $\frac{-1}{4} \leq ab \leq \frac{4}{9}$. Solution of Zhangyanzong: $$2(1-c^2)=2(a^2+b^2)\geq (a+b)^2=(1-c)^2$$$$3c^2-2c-1\leq 0,-\frac{1}{3}\leq c\leq 1$$$$ab=c^2-c=(c-\frac{1}{2})^2-\frac{1}{4}\geq -\frac{1}{4}$$$$ab=c^2-c=(c-\frac{1}{2})^2-\frac{1}{4}\leq (-\frac{1}{3}-\frac{1}{2})^2-\frac{1}{4}=(-\frac{1}{3})^2-(-\frac{1}{3})= \frac{4}{9}$$

From $1-c^2+2ab=a^2+b^2+2ab=(1-c)^2$, we have $ab=c(c-1).$ Then, \[(c-\frac{1}{2})^2\ge 0 \iff c(c-1)\ge -\frac{1}{4}.\]Next, \[c(c-1)\le \frac{4}{9} \iff (3c+1)(3c-4)\le 0\]since $|c|\le 1,$ assume by contradiction that $c<-\frac{1}{3},$ then \[a^2+b^2=1-c^2 <\frac{8}{9}\]\[\frac{8}{9}+2ab>a^2+b^2+2ab=(1-c)^2>\frac{16}{9} \implies ab>\frac{4}{9}\]this gives \[\frac{8}{9}>a^2+b^2\ge 2ab >\frac{8}{9}\]a contradiction. Therefore, $c\ge -\frac{1}{3}$ and $c(c-1)\le \frac{4}{9}.$

nice one