Proving $2\left(b^2+c^2\right)-a^2\leqslant 12$ with some condition. Problem. Let $a,b,c\in\mathbb{R}$ such that $a+b+c=6,$ $a^2+b^2+c^2\in\left[12,\frac{68}3\right]$ and $a\geq b\geq c.$ Prove
$$2\left(b^2+c^2\right)-a^2\leqslant 12.$$
When do we have equality?
I can only prove it for $a,b,c \geqslant 0.$
From $a+b+c=6$ we have $a=6-b-c \geqslant 0.$ We need to prove the inequality when $$\Big[6-b-c\geqslant 0, \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12\geqslant 0,\\{\frac {68}{3}}-
 \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2}\geqslant 0,6-c-2\,b\geqslant 0,b-c\geqslant 0\Big]$$
By computer we have$:$ $$12+ \left( 6-b-c \right) ^{2}-2\,{b}^{2}-2\,{c}^{2}$$
$$= \left[  \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right]  \Big[\left( 
b-c \right)  \left( {\dfrac {5}{32}}\,{c}^{2}+\dfrac{1}{16}\,bc \right)  +\left( 
6-c-2\,b \right)  \left( {\frac {3}{64}}\,{b}^{2}+\dfrac{1}{4}b+{\frac {5}{32
}}\,bc \right) \Big]
$$
$$+{\frac {3}{32}} \left[ {\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b
}^{2}-{c}^{2} \right] \left( b-c \right) {c}^{2}+\dfrac{1}{2} \left( 6-b-c
 \right)  \left( 6-c-2\,b \right) c$$
$$+\dfrac{1}{6}\, \left( 6-c-2\,b \right) ^{2}
{b}^{2}+ \left(  \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right) ^{2}
 \left( \dfrac{1}{12}+{\frac {3}{256}}\,{c}^{2}+\dfrac{1}{32}c+{\frac {3}{128}}\,bc
 \right) +\dfrac{1}{6} \left( 6-c-2\,b \right)  \left( b-c \right) bc+\dfrac{1}{6}\,
 \left( b-c \right) ^{2}{c}^{2}$$
$$+ \left[  \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right]  \left[
{\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2} \right]
 \left( {\frac {3}{256}}\,{c}^{2}+{\frac {3}{128}}\,bc \right) $$
$$+ \left[ {\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2}
 \right]  \left( 6-c-2\,b \right)  \left( {\frac {3}{64}}\,{b}^{2}+{
\frac {3}{32}}\,bc \right)\geqslant 0 $$
For the text of the above decomposition, please see in my file: Click here.

Let $s = b - c \ge 0$. From $a = 6 - b - c \ge b$, we have $c \le 2 - \frac{2s}{3}$.
We have $a^2 + b^2 + c^2 = (6-b-c)^2 + b^2 + c^2 = 6(c - 2 + s/2)^2 + s^2/2 + 12$.
From $a^2 + b^2 + c^2 \in [12, 68/3]$, we have
$$6(c - 2 + s/2)^2 + s^2/2 + 12 \le \frac{68}{3}$$
and thus
$$0 \le s \le \frac{8}{\sqrt{3}}, \quad 2 - s/2 - \frac{1}{6}\sqrt{64-3s^2}
\le c \le 2 - s/2 + \frac{1}{6}\sqrt{64-3s^2}.$$
Clearly, $2 - \frac{2s}{3} \le 2 - s/2 + \frac{1}{6}\sqrt{64-3s^2}$
for $0 \le s \le \frac{8}{\sqrt{3}}$.
From $2 - s/2 - \frac{1}{6}\sqrt{64-3s^2} \le 2 - \frac{2s}{3}$,
we have $0 \le s \le 4$.
Then, we have $2(b^2+c^2) - a^2 = s^2+24c+12s-36 
\le s^2 + 24(2 - \frac{2s}{3}) + 12s - 36 = 12 - s(4-s) \le 12$.
We are done.
