<s>[INST] 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: $\sqrt{6}<b\le\sqrt{7}$
 [/INST] For the lower bound we have: $b > |a-c| \implies b^2 > (a-c)^2, (2b)^2 = (a+c)^2\implies 5b^2 > (a-c)^2 + (a+c)^2 = 2(a^2+c^2) = 2(21- b^2) \implies 7b^2 > 42 \implies b^2 > 6 \implies b > \sqrt{6}$.
</s>