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 $.