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!

Rewriting the inequality as  :
$$\frac{1}{2}\left(-bc+\sqrt{b^2c^2-4\left(b+c\right)^{2}+8bc+16}\right)+b+c \geq \dfrac{3+\sqrt{7}}{2}$$
We make :
$$b+c=\operatorname{constant}=C$$
The function :
$$g(bc)=-bc+\sqrt{b^2c^2-4\left(C\right)^{2}+8bc+16}$$
Is increasing as $bc$ increases with $-bc+\sqrt{b^2c^2-4\left(C\right)^{2}+8bc+16}=a\geq 0$
So if $b=0$ with the constraint we have an equality as $c=1.5$
Done .
