find the range of the $b$, if $a^2+b^2+c^2=21~~~~2b=a+c$ Let $\Delta ABC$ such $|AC|=b,|BC|=a,|AB|=c$,and such $$\begin{cases}
2b=a+c\\
a^2+b^2+c^2=21
\end{cases}$$
find the range of the $b$
since
$$(a+c)^2-2ac+b^2=21$$
so we have
$$2ac=5b^2-21$$
then $a,c$ is a  equation
$$x^2-2bx+\dfrac{5b^2-21}{2}=0$$ roots.so we have
$$\Delta =4b^2-2(5b^2-21)=42-6b^2\ge 0\Longrightarrow -\sqrt{7}\le b\le \sqrt{7}$$
the book  aswer: $\sqrt{6}<b\le\sqrt{7}$

Upper bound
Since $a+c=2b$ we have that $a+b+c=3b.$ Now, squaring we get
$$9b^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc.$$ Using that $a^2+b^2+c^2=21$ and $ab+ac+bc\le a^2+b^2+c^2=21$ we get $b^2\le 7.$ So, we have that upper bound $b\le \sqrt{7}.$
