Schureal! Topic: Inequalities. Level: Olympiad.

Problem: (WOOT Message Board, Mildorf) Let $ a,b,c $ be arbitrary reals such that $ a \ge b \ge c $, and let $ x, y, z $ be nonnegative reals such that $ x+z \ge y $. Prove that

$ x^2(a-b)^k(a-c)^k+y^2(b-c)^k(b-a)^k+z^2(c-a)^k(c-b)^k \ge 0 $

where $ k $ is a positive integer.

Solution: Let's make a substitution which will help make things look a little nicer. Define $ a-b = d $ and $ b-c = e $, so $ a-c = d+e $ with $ d,e $ nonnegative reals.

For the case $ k = 1 $, we want to show that

$ x^2d(d+e)+z^2e(d+e) \ge y^2de $.

From the given condition, we have $ (x+z)^2de \ge y^2de $. (1)

Also, we know $ (xd-ze)^2 \ge 0 $. (2)

Adding up (1) and (2), we have

$ x^2d(d+e)+z^2e(d+e) \ge y^2de $

as desired.

Now we will induct on this base case. Assume for some positive integer $ n $ we have

$ x^2d^n(d+e)^n+z^2e^n(d+e)^n \ge y^2d^ne^n $. (3)

We clearly have

$ x^2d^{n+1}(d+e)^{n+1}+z^2e^{n+1}(d+e)^{n+1} = x^2d^n(d+e)^n \cdot [d(d+e)]+z^2e^n(d+e)^n \cdot [e(d+e)] $.

But since $ d(d+e) \ge de $ and $ e(d+e) \ge de $, we know that the above expression is at least

$ x^2d^n(d+e)^n \cdot (de)+z^2e^n(d+e)^n \cdot (de) \ge y^2d^ne^n (de) = y^2d^{n+1}e^{n+1} $

by (3), thus completing the induction. So the given inequality is true for all positive integers $ k $. QED.

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Comment: This is very similar to Schur's Inequality with more variables and some interesting conditions. The $ d, e $ substitution is very useful as it allows us to restrict ourselves to nonnegative reals, making multiplication on both sides of an inequality valid. We also had to find a good way to use the $ x+z \ge y $ condition, which is seen in the solution.

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Practice Problem: