If $a+b+c=1$, find the minimum of $\frac{4+3abc}{ab+bc+ac}$ I came to ask this because I am really stuck at this problem. I have tried everything from arithmetic mean, geometric mean and harmonic mean. Also, I have tried playing with the variables and such, but it got me to nowhere.
If $a+b+c=1$; $a,b,c$ nonnegative, calculate the minimum of 
$$\frac{4+3abc}{ab+bc+ac}$$
All I've got so far is:
$$\frac{3abc}{ab+bc+ac} \le \frac{1}{3}$$
But this is obviously on the wrong side of the inequality.
Also, I think that
$$\frac{1}{ab+bc+ac}\ge3$$
But I haven't been able to prove it.
Playing with the most possible and obvious values, one could think that the answer is 37/3, but the excercise is about proving it. Any help and little hints are greatly apprecieated.

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