**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 $.

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

ReplyDeleteWe 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.

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

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

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.

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

ReplyDelete