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

[anemone]

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.

 

[Greg]

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.

Spoiler

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 $f(x)=\left(x-\dfrac12\right)^3$.

 

[anemone]

Spoiler

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!

 

[anemone]

Hint:

Spoiler

Schur's inequality.

 

[anemone]

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:

Spoiler

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. $f(x)=\left(x-\dfrac{1}{2}\right)^3$.