Absolutely less than Quadratic
Let a , b , c be real numbers such that
∣ a x + b ∣ ≤ ( x + c ) 2 for all x ∈ R .
Find the maximum value of a b + b c + c a .
The answer is 0.
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3 solutions
Check the inequality at x = − c ,
∣ b − a c ∣ ≤ 0
Now since absolute function cannot be negative, ∣ b − a c ∣ = 0 ⇒ b = a c .
Now the derivative of ( x + c ) 2 is 0 at x = − c , so the derivative of ∣ a x + b ∣ at x = − c + and x = − c − must be 0. Therefore, a = 0 .
Substituting in a b + b c + c a , we get 0.
Point to note: The derivative of ∣ a x + b ∣ does not exist at a x + b = 0 if a = 0 . So, in a sense, you already assumed that a = 0 , to conclude that a = 0 .
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Is it fine now? I had the graph of functions in my mind while writing the solution.
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It is far better to explicitly talk about the graphical implication.
This is somewhat better, but then the statement is not true, because the implication only works one direction, which is not the direction that you want. It is not that the derivative must be 0, but rather that the derivative is at least / at most 0 (depending on which side). You might run into issues justifying this analytically, using what we know about polynomials and their behavior.
Clearly , LHS is a equation of line and RHS is equation of a parabola opening upwards. The inequality is valid when line is tangent to parabola at its lowest possible value i.e. y=0 (X-axis). Hence, a=0, b=0. Hence ab+bc+ca = 0.
A purely algebraic solution :
x = − c implies b = a c .
x = 2 a − c implies a = 0 .
So b = 0 and a b + b c + c a = 0 .