MHB What Is the Cubic Polynomial f(x) When 16a + 24b = 9?

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The discussion revolves around determining the cubic polynomial f(x) = x^3 - (3/2)x^2 + ax + b, with real roots constrained to the interval (0, 1). Participants are tasked with proving that 16a + 24b ≤ 9 and finding the specific function when equality is achieved. The problem invites alternative approaches to reach the solution, indicating a collaborative problem-solving environment. The emphasis is on mathematical proofs and exploring the implications of the polynomial's coefficients. Ultimately, the goal is to establish the conditions under which the polynomial meets the specified criteria.
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Let the function of $f$ be a cubic polynomial such that $f(x)=x^3-\frac{3}{2}x^2+ax+b=0$, with real roots lie in the interval $(0,\,1)$.

Prove that $16a+24b\le 9$. Find the corresponding function of $f$ when the equality holds.
 
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anemone said:
Let the function of $f$ be a cubic polynomial such that $f(x)=x^3-\frac{3}{2}x^2+ax+b=0$, with real roots lie in the interval $(0,\,1)$.

Prove that $16a+24b\le 9$. Find the corresponding function of $f$ when the equality holds.

For ease of notation I will use $A$ as $a$ and $B$ as $b$.

$$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$$
$$\Rightarrow A=ab+ac+bc,B=-abc,a+b+c=\dfrac32$$

$$(a+b+c)^2=a^2+b^2+c^2+2A=\dfrac94\Rightarrow a^2+b^2+c^2=\dfrac94-2A\quad(1)$$

$$(a-b)^2\ge0$$

$$a^2+b^2\ge2ab$$

Similarily,

$$a^2+c^2\ge2ac$$

$$b^2+c^2\ge2bc$$

Adding and simplifying gives

$$a^2+b^2+c^2\ge A$$

From $(1)$:

$$\dfrac94-2A\ge A$$

$$\dfrac34\ge A\quad(3)$$

From the AM-GM inequality:

$$\dfrac{a+b+c}{3}\ge\sqrt[3]{abc}\Rightarrow-\dfrac18\le B\quad(4)$$

From $(3)$ and $(4)$

$$12\ge16A$$

$$3\ge-24B$$

$\Rightarrow9\ge16A+24B$ with equality when $a=b=c=\dfrac12$.

The corresponding function for $f$ is $f(x)=\left(x-\dfrac12\right)^3$.
 
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greg1313 said:
For ease of notation I will use $A$ as $a$ and $B$ as $b$.

$$(x-a)(x-b)(x-c)=x^3-\dfrac32x^2+(ab+ac+bc)x-abc$$
$$\Rightarrow A=ab+ac+bc,B=-abc$$

$$(a+b+c)^2=a^2+b^2+c^2+2A=\dfrac94\Rightarrow a^2+b^2+c^2=\dfrac94-2A\quad(1)$$

$$(a-b)^2\ge0$$

$$a^2+b^2\ge2ab$$

Similarily,

$$a^2+c^2\ge2ac$$

$$b^2+c^2\ge2bc$$

Adding and simplifying gives

$$a^2+b^2+c^2\ge A$$

From $(1)$:

$$\dfrac94-2A\ge A$$

$$\dfrac34\ge A\quad(3)$$

From the AM-GM inequality:

$$\dfrac{a+b+c}{3}\ge\sqrt[3]{abc}\Rightarrow-\dfrac18\le B\quad(4)$$

From $(3)$ and $(4)$

$$12\ge16A$$

$$3\ge-24B$$

$\Rightarrow9\ge16A+24B$ with equality when $a=b=c=\dfrac12$.

The corresponding function for $f$ is $f(x)=\left(x-\dfrac12\right)^3$.

Awesome, greg1313! And thanks for participating!(Cool)

This problem can still be solved using another route, and I welcome those who are interested to take a stab at it!
 
Hint:

Schur's inequality.
 
anemone said:
Let the function of $f$ be a cubic polynomial such that $f(x)=x^3-\frac{3}{2}x^2+ax+b=0$, with real roots lie in the interval $(0,\,1)$.

Prove that $16a+24b\le 9$. Find the corresponding function of $f$ when the equality holds.

Solution of other:

Let the three roots be $p,\,q$ and $r$. By Schur's inequality, we have:

$(p+q+r)^3+9pqr\ge 4(p+q+r)(pq+qr+rp)$

which is just

$\left(\dfrac{3}{2}\right)^3+9(-b)\ge 4\left(\dfrac{3}{2}\right)\left(a\right)$

and upon simplification we get:

$16a+24b\le 9$

Equality holds when $p=q=r$, i.e. $f(x)=\left(x-\dfrac{1}{2}\right)^3$.
 
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