If $a+b+c=1$, find the minimum of $\frac{4+3abc}{ab+bc+ac}$

It seems you are right and the minimum is indeed 37/3. We shall use the standard techniques for proving the inequalities.

If we put $a=b=c=1/3$ then the obtain the upper bound $37/3$ for the minimum. It rests to show that holds

(1) $\frac{4+3abc}{ab+bc+ac}\ge 37/3.$

At first we homogenize [Lee, Ch 3] the left side

$$\frac{4+3abc}{ab+bc+ac}=\frac{4(a+b+c)^3+3abc}{(ab+bc+ac)(a+b+c)}.$$

Expanding and simplifying, we reduce inequality (1) to the form

$$12(a^3+b^3+c^3)\ge (a^2b+ab^2+ab^2+ac^2+b^2c+bc^2)+30abc,$$

which should follow from Muirhead Theorem [Lee, Ch. 3.3].

References

[Lee] Hojoo Lee. Topics in Inequalities - Theorems and Techniques (February 25, 2006).

Substitute $a=1-(b+c)$ in $\displaystyle\frac{4+3abc}{ab+bc+ac}$ to get

$$\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c} \tag{1}$$

Differentiating $(1)$ with respect to $b$ gives

$$\frac{\partial}{\partial b}\left(\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}\right) =\frac{(3c^3-3c^2+4)(2b-c+1)}{\left(b^2+b(c-1)+c(c-1)\right)^2}$$

Setting this equal to zero yields

$$2b-c+1=0 \tag{2}$$

Differentiating $(1)$ with respect to $c$ gives

$$\frac{\partial}{\partial c}\left(\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}\right) =\frac{(3b^3-3b^2+4)(2c-b+1)}{\left(c^2+c(b-1)+b(b-1)\right)^2}$$

Setting this equal to zero yields

$$2c-b+1=0 \tag{3}$$

Solving for $(2)$ and $(3)$ yields

$$b=c=\frac{1}{3}$$

Plug these values in $(1)$ and you end up with

$$\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}=\frac{37}{3}$$