Here is my take on this question, I could be wrong (inequalities are tricky for me).
From the statement f(0)=81, we have d=81
This, f(x) becomes 3x^3+bx^2+cx+81.
Let the roots be r, s, and t. Thus, the product rst is equal to -27 by Vieta's formulas.
b=-(r+s+t), which is a positive number, given that r, s, and t are all negative. Hence, to minimize b, we want to maximize r+s+t, knowing that they are negative. We seek to relate rst to r+s+t through inequality, and the AM-GM works for this. The problem arises though, that the AM-GM only works for positive numbers, and r, s, and t are each negative. Luckily, by setting -r, -s, and -t into the equation, we have all positive inputs, hence allowing for the AM-GM to come into use.
The AM-GM for 3 variables gives \(\frac{-r-s-t}{3}≥ \sqrt[3]{-rst}\), simplifying to \(-(r+s+t)≥ 3\sqrt[3]{27}\) or \(r+s+t≤-9\) from this, the maximum value of r+s+t is -9, so the minimum value for b is 9. This can only be achieved when r=s=t however, so r=s=t=-3
Now to find the minimum value for c
With r, s, and t each being negative rs, st, and rt must all be positive. From this, it is easy to apply AM-GM for 3 variables
\(\frac{rs+st+rt}{3}≥\sqrt[3]{rst^2}\), simplifying to \(rs+st+rt≥27\). Equality only occurs when the equality condition of AM=GM is met, however, or when rs=st=rt, r=s=t. From this, the minimum value of c is 27.
Combining the minimum value of b with the minimum value of c, we find that both can be achieved only when r=s=t=-3. Through this, the roots must be -3, -3, and -3, with b=9, c=27. F(1) is equal to 3+b+c+81, or 3+9+27+81, or 120.
Hopefully, this is a valid solution, have a great day!
Note: After expanding (x+3)^3, and multiplying by 3, I found that this solution is incorrect. Hopefully, you can find inspirational value from this then. Where I went wrong was when I tried to use AM-GM to find b, as the application was completely out of the restriction of the AM-GM.
Someone previously asked this question on this website though, and the answer they got was 204.
