Prove $a+b+c \geq \frac{3+\sqrt{7}}{2}$ for $a,b,c \in [0, \frac32]$ with $a^2+b^2+c^2+abc=4$ 
Let $a,b,c \in \left[0;\frac{3}{2}\right]$ and $a^2+b^2+c^2+abc=4$. Prove that
$$a+b+c \geq \dfrac{3+\sqrt{7}}{2}.$$

Source: This is a math problem that my teacher gave me $3$ months ago (the submission deadline has expired). My teacher wrote a book and sent me to test the difficulty of the problem.
My attempt: I have converted $a$ to $b,c$, used to trigonometric conversion, but all failed.
Related problem (the same source, with the same conditions): https://artofproblemsolving.com/community/c6h2975620p26673082
Please give me a suggestion! Thank you!

It suffices to prove that
$$2a + 2b + 2c - 3 \ge \sqrt 7.$$
Using $(a + b + c)^2 \ge a^2 + b^2 + c^2$ and $(a + b + c)^3/27 \ge abc$, we have
$(a + b + c)^2 + (a + b + c)^3/27 \ge 4$
which results in $a + b + c > 3/2$ or
$2a + 2b + 2c - 3 > 0$.
Thus, it suffices to prove that
$$(2a + 2b + 2c - 3)^2 - 7 \ge 0.$$
We have
\begin{align*}
 &(2a + 2b + 2c - 3)^2 - 7 - 4(a^2 + b^2 + c^2 + abc - 4)\\
 ={}& (-4ab + 8a + 8b - 12)c + 2(3-2a)(3-2b)\\
 \ge{}& 0. \tag{1}
\end{align*}
(Note: If $-4ab + 8a + 8b - 12 \ge 0 $, clearly (1) is true.
If $-4ab + 8a + 8b - 12 < 0$, we have
$(-4ab + 8a + 8b - 12)c + 2(3-2a)(3-2b)$
$\ge (-4ab + 8a + 8b - 12)\cdot \frac32 + 2(3-2a)(3-2b)$
$ = 2ab \ge 0$.)
We are done.
