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.