## Sunday, May 28, 2006

### Don't Mess With The Juggernaut. Topic: Inequalities. Level: Olympiad.

Problem: Given nonnegative reals $a,b,c$ satisfying $a+b+c = 3$, find the minimum of the expression

$a^3+b^3+c^3+kabc$

for $k \ge \frac{15}{4}$.

Solution: At first glance, we're like it's just a boring classical homogeneous symmetric inequality. The minimum must occur at $a = b = c = 1$! But looking into it a little more, we find that if $a = b = \frac{3}{2}, c = 0$ we get something smaller. In fact, we will show that this is the minimum. We claim that

$a^3+b^3+c^3+kabc \ge \frac{(a+b+c)^3}{4}$.

Expanding and rearranging, this is equivalent to

$3(a^3+b^3+c^3)+(4k-6)abc \ge 3(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)$.

But wait, this form reminds us a lot of Schur! Schur says that (multiplied by a factor of $3$)

$3(a^3+b^3+c^3)+9abc \ge 3(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)$.

So it remains to show that

$(4k-15)abc \ge 0$.

But since $4k-15 \ge 0$ (from $k \ge \frac{15}{4}$) and $a,b,c$ are nonnegative, the inequality is clearly true. Hence we have

$a^3+b^3+c^3+kabc \ge \frac{(a+b+c)^3}{4} = \frac{27}{4}$

for $k \ge \frac{15}{4}$. Note that the equality conditions are met in both inequalities when $a = b = \frac{3}{2}$ and $c = 0$. QED.

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Comment: Original problem was motivated by the case $k = 8$, but easily generalized by seeing the proof. This is an interesting inequality because it doesn't have the usual equality case except when $k = \frac{15}{4}$.

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Practice Problem: Solve for $x \in \mathbb{R}$

$2x^2-3x\lfloor x-1 \rfloor+\lfloor x-1 \rfloor^2 \le 0$.

1. (Wow, when I first saw that problem I thought it was hard.)

We can factor!

(2x - \lfloor x-1 \rfloor)(x - \lfloor x-1 \rfloor) \le 0

Let u = \lfloor x-1 \rfloor and let d = fpart(x-1). Then

(2(u + d + 1) - u)(u + d + 1 - u) \le 0
(u + 2d + 1)(d + 1) \le 0

The second factor is always positive, so the first must be negative. The first solution comes at x = -1.

For -2 \le x < -1 we have u = -2 and hence we want (2d - 1) \le 0. This requires d \le 1/2, and o ur second solution set comes at -2 \le x \le -1.5.

For x < -2 we have u = -3 and the first factor will always be negative.

QED.

2. Oh snap the first factor should be (u + 2d + 2). And for -2 \le x < -1 we have u = -3, so my mistakes cancel out and the solutions are the same XD

Alternately, \lfloor x-1 \rfloor = \lfloor x \rfloor - 1, which simplifies the problem a bit and makes it harder to make dumb mistakes.

3. Yeah, another way to do it is to set a = x, b = [[x-1]] and factor that. It was on AoPS, but they missed the negative solutions.

4. [...] have the minimum result generalized here. The proof is the same [...]