# Prove that $0 \leq ab^2-ba^2 \leq \frac{1}{4}$ with $0 \leq a \leq b \leq 1$.

Let $a$ and $b$ be real numbers such that $0 \leq a \leq b \leq 1$. Prove that $0 \leq ab^2-ba^2 \leq \dfrac{1}{4}$.

Attempt

We can see that $ab^2-ba^2 = ab(b-a)$, so it is obvious that it is greater than or equal to $0$. But how do I show it is also less than or equal to $\dfrac{1}{4}$?

• Try having$1/2$ as the defining line, you get 3 cases, either both are less than one half, both greater than one half, or only one is greater and only one is less that (or equal in all cases) and prove each one. – Zelos Malum Dec 23 '15 at 18:39

Hint: write $b=a+c$. Then $ab(b-a)=ac(a+c)$. Now apply AM-GM inequality to $a,c$ and recall $a+c=b\leq 1$.

Elaboration on AM-GM part: $\sqrt{ac}\leq\frac{a+c}{2}\leq\frac{1}{2}$ so $ac\leq\frac{1}{4}$.

• Can you elaborate on the AM-GM part? – user19405892 Dec 23 '15 at 18:43
• Here you go. ${}$ – Wojowu Dec 23 '15 at 18:44

First proof

The part $0 \leq ab^2-ba^2$ is equivalent with $$0 \leq ab(b-a)$$ which is true because $b \geq a \geq 0$ .

For the other part use $b^2 \leq b \leq 1$ and $a^2 \leq a$ as follows :

$$ab^2-ba^2 \leq ab-ba^2=b(a-a^2) \leq a-a^2 \leq \frac{1}{4}$$ the last part being equivalent with $\left (a-\frac{1}{2} \right )^2 \geq 0$ .

Second proof

Denote $f(a,b)=ab^2-ba^2=ab(b-a)$ . Now notice that if $d>0$ then :

$$f(a+d,b+d)=(a+d)(b+d)(b-a) \geq ab(b-a)=f(a,b)$$

In this way we can increase $b$ to $1$ by choosing $d=1-b$ (this technique is usually called smoothing ). So it suffices to prove that :

$f(a+d,1) \leq \frac{1}{4}$ which is as in the previous proof equivalent with $$\left (a+d -\frac{1}{2} \right )^2 \geq 0$$

• if $b<1$ then $b^2<b$ – Leox Dec 23 '15 at 19:08
• @Leox Thanks , I wrote it backwards but used it right . – user252450 Dec 23 '15 at 19:09

There are more generic ways of solving such problems. Since $f(a,b) = ab^2 - ba^2$ is continuous and differentiable, and the domain $0 \le a \le b \le 1$ is a compact simplex, it attains its max and min at one of the following points:

1. A point where $0 < a < b < 1$ and $\nabla f (a,b) = 0$, i.e. $f_x(a,b) = f_y(a,b) = 0$

2. A point where $a = 0$, $0 < b < 1$, and $f_y(a,b) = 0$

3. A point where $0 < a < 1$, $b = 1$, and $f_x(a,b) = 0$

4. A point where $0 < a = b < 1$, and $f_x(a,b) + f_y(a,b) = 0$

5. A point where $a = b = 0$ or $a = b = 1$.

Calculate that $f_x(a,b) = b^2 - 2ab$, $f_y(a,b) = 2ab - a^2$. There are no points of type (1). Type (2) points are anything on the line, and here $a = 0$ so $f(a,b) = 0$. Type (3) points satisfy $b = 1$, so $1 - 2a = 0$, so $a = \frac12$, and here $f(a,b) = \frac12 - \frac14 = \frac14$. Type (4) points all have $f(a,b) = 0$. Finally, the two type (5) points are $f(0,0) = 0$ and $f(1,1) = 0$.

Thus the minimum of $f$ is $0$ and the maximum is $\frac14$, on this domain.