Maximize This

Relevant wiki: Smoothing an Inequality

Note that without loss of generality, we may assume $a \ge b \ge c$ . This would imply that $3a \ge 3 \ge 3c (\because a+b+c=3)$ .

So we have $a \ge 1 \ge c$ . Since $1 \ge c \ge 0$ , we have that $c^2-c+1 \le 1$ .

Also, since $f(x)=x^2-x+1$ is a increasing function for $[1, \infty)$ we know that $f(a+c) f(0) \ge f(a)f(c)$ .

This would imply that $\prod_{cyc} (a^2-a+1) \le f(a+c)f(b)f(0)=f(b)f(3-b)\times 1=(b^2-b+1)(b^2-5b+7)$ .

Note $b \in [0, \frac{3}{2}]$ , and that the derivative of the above expression is $(2b-3)(b-1)(b-2)$ .

This would imply that when $b \in [0, 1)$ the function is decreasing, but when $b \in (1, \frac{3}{2}]$ the function is increasing.

The maxima thus must be when either $b=0$ or $b=\frac{3}{2}$ , but the value when $b=0$ (which is $7$ ) is larger. Thus the answer is $7$ .

I do not know whether this will be accepted as a formal solution, but lemme try...

Assume WLOG that $a \geq b \geq c$ . Then since we have $a+b+c = 3$ , then we have to have $c \in [0,1]$ . Now, to maximize $c^2-c+1$ for $c \in [0,1]$ , we must have either $c = 1$ or $c = 0$ . If $c = 1$ then the expression evaluates to $1$ , which is probably not the maximum. So, we have to set $c = 0$ so that the expression becomes $(a^2-a+1)(b^2-b+1)$ with $a+b = 3$ .

Now, we can let $b = 3-a$ so that we only need to maximize $((a^2-a+1)((3-a)^2-(3-a)+1)$ . Differentiating and equating to 0, we see that a local maxima is found when $a = b = 1.5$ , giving us a value of $3.063$ . But then, since the quartic function is increasing as one moves to the right of the 3rd local minima, and the third local minima is found at $a = 2$ , then we need to consider when $a$ is maximized and greater than $2$ , i,e, $a = 3, b = 0$ . Since this yields a value of $7$ for the function, one can say that the maximum value is $\boxed{7}$ .

Let $f(x)=\ln(x^2-x+1)$ and $g(x)=\dfrac{\ln(7)}{3} x$ .

Lemma: For all $0\le x \le 3$ we have $f(x)\le g(x)$ .

Proof of Lemma:

$f'(x) = \dfrac{2x-1}{x^2-x+1}$ and $f''(x)= \dfrac{-2x^2+2x+1}{(x^2-x+1)^2}$ .

Let $k= \dfrac{1+\sqrt{3}}{2}$ . Thus $f''(x) > 0$ for $0\le x < k$ , and $f''(x) <0$ for $x > k$ .

Now $f(0)=0$ and $f(k)= \ln\left(\dfrac{3}{2}\right)$ . Therefore, for $0\le x\le k$ we have: $f(x) \le \dfrac{f(k)}{k} x = \ln\left(\dfrac{3}{2}\right)\left(\sqrt{3}-1\right) x \le \dfrac{\ln(7)}{3} x = g(x)$

Also we have $f'(3)= \dfrac{5}{7} > \dfrac{\ln(7)}{3}$ . Thus for all $k < x \le 3$ , we have $f'(x) \ge f'(3) > g'(x)$ and $f(3) = g(3)$ . Thus by the fundamental theorem of calculus, we have $f(x) \le g(x)$ for $k < x \le 3$ .

Therefore the proof of the Lemma is complete. $\square$

Solution of the problem:

Note that since $a+b+c=3$ and $a,b,c \ge 0$ , then each of $a,b,c \le 3$ .

Thus, the Lemma implies that $x^2-x+1 \le 7^{x/3}$ for $x = a,b,c$ .

Therefore, $(a^2-a+1)(b^2-b+1)(c^2-c+1) \le 7^{(a+b+c)/3} = \boxed{7}$

This value is achieved when $a=0, b = 0, c=3$ .

I used spatial imagination to solve this. Visualize the equation $y=x^2-x+1$ You can see $y=1$ for both $x=0$ and $x=1$ and the minimum occurs at $x=0.5$ . Now, imagine there are 3 such graphs and each graph has a cursor at $x=1$ . The multiplication of these graphs gives us the main expression. You can move the cursor along the x-axis of each graph. The reading the cursors give can be labelled as $a, b, c$ . The problem gave us a restriction which is $a+b+c=3$ . Now, move the 'a' cursor to the right. For the restriction to be satisfied, another cursor will move equal distance to the left. Let that cursor be the 'b' cursor. You can see the most logical situation would be to let the 'b' cursor to get to the point $x=0$ as it'll give the same value as its initial position $x=1$ gave. Simultaneously the 'a' cursor has moved to the point $x=2$ . Hence, the function we're working on increased in the value as well. After that, we'll try to move the 'a' cursor to the right again. This time the 'c' cursor has to move to the left. Like the 'b' cursor, the 'c' cursor will also stop at $x=0$ . Meanwhile, the 'a' cursor will stop at $x=3$ maintaining the restriction. Therefore, we've got $a=3, b=0, c=0$ which corresponds to the maximum value of the given function. Now, putting these values into the expression, we find the maximum value must be 7.

Since all the three brackets basically just contain the same equation, it doesn't matter which variable is assigned what value to. So, there's following ways to get a + b + c = 3

• (a, b, c) = (1, 1, 1)
• (a, b, c) = (0, 1, 2)
• (a, b, c) = (0, 0, 3)

It's easy to see that the last combination (0, 0, 3) propels the given problem statement to the highest value of 7