## Monday, March 13, 2006

### 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: