All of the roots of the cubic polynomial \(f(x) = 3x^3+bx^2+cx+d \) are negative real numbers. If f(0) =81, find the smallest possible value of f(1). Prove that your answer is the minimum.

 Jun 4, 2021

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.  

 Jun 4, 2021
edited by EnchantedLava68  Jun 4, 2021

After reviewing my work, the next day after this question I found my mistake was not multiplying b and c by 3.  This was because f(x) had roots of -3, -3, -3, but it was multiplied by 3, hence giving -b/3=-9, and c/=27


From this, b=27 and c=81, so f(1) = 3+27+81+81 = 162+30=192

EnchantedLava68  Jun 5, 2021

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