MHB Find Lowest Value for A: a1, a2, a3 & 4 | Arithmetic Progression

[mitaka90]

a₁, a₂, a₃ and 4 make an arithmetic progression with difference d. For which values of d, A = a₁a₂ + a₂a₃ + a₃a₁ has the lowest value?I don't know if I went with the right approach, but I managed to get this : A=3x² +6xd + 2d² for a₁= x, a₂ = x + d, etc... But I don't know what else to do.

 

[anemone]

a₁, a₂, a₃ and 4 make an arithmetic progression with difference d. For which values of d, A = a₁a₂ + a₂a₃ + a₃a₁ has the lowest value?I don't know if I went with the right approach, but I managed to get this : A=3x² +6xd + 2d² for a₁= x, a₂ = x + d, etc... But I don't know what else to do.

Hi mitaka90!

It seems to me you haven't made good use of the given fourth term in that arithmetic progression...:) the fourth term would help you to simplify your $A$ in terms of only one variable and when you have the quadratic equation in terms of one variable, I believe you could handle from there...

 

[mitaka90]

Hi mitaka90!

It seems to me you haven't made good use of the given fourth term in that arithmetic progression...:) the fourth term would help you to simplify your $A$ in terms of only one variable and when you have the quadratic equation in terms of one variable, I believe you could handle from there...

Omg, I'm such a moron. I hate it when I do the hard work and then the easiest and most noticable thing just slips from my sight. Thank you sincerely, I guess that little tip is what I needed.