ISL 2005 A3

by Wolstenholme, Sep 6, 2014, 2:25 AM

Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.

Proof:

Assume WLOG that $ p \ge q \ge r \ge s $. Note that $ pq + pr + ps + qr + qs + rs = \frac{(p + q + r + s)^2 - (p^2 + q^2 + r^2 + s^2)}{2} = 30 $. By the rearrangement inequality, we have that $ pq + rs = \max(pq + rs, pr + qs, ps + qr) $ and so $ pq + rs \ge 10 $. Letting $ x = p + q $ we have that $ x^2 + (x - 9)^2 = 21 + 2(pq + rs) \ge 41 $ which becomes $ (x - 4)(x - 5) \ge 0 \Longrightarrow x \ge 5 $. This means that $ 2\sqrt{rs} \le r + s \le 4 \Longrightarrow rs \le 4 $. But since $ pq + rs \ge 10 $ this means that $ pq \ge 6 $ and so we are done.

Now I want to discuss motivation. After some playing around we see the only equality case is $ (p, q, r, s) = (3, 2, 2, 2) $. Now, looking at this, we want the extreme values of the sum of the two biggest or the two smallest to be greater than $ 5 $ and less than $ 4 $ respectively, so letting $ x = p + q $ we want to get exactly the equation $ (x - 4)(x - 5) \ge 0 $. Now we want an $ x - 9 $ in there somewhere, and rearranging it turns out we want $ x^2 + (x - 9)^2 \ge 41. $ But this is the same as $ pq + rs \ge 10 $, and when looking at a sum like this, the rearrangement inequality should come to mind immediately.

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  • Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial :P

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