Olympiad Maths Prep

Track / Stage 7 / 228 of 300 #1628 of 2000

Problem 1628

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.5 Prove it

47. Let a1,a2,,an;b1,b2,,bna_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n} be two sequences of positive numbers. Prove: (ijaibj)2\left(\sum_{i \neq j} a_{i} b_{j}\right)^{2} \geqslant (ijaiaj)(ijbibj)(1998\left(\sum_{i \neq j} a_{i} a_{j}\right)\left(\sum_{i \neq j} b_{i} b_{j}\right) \cdot(1998 Yugoslav Mathematical Olympiad problem)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

47. Let A=i=1nai,B=i=1nbiA=\sum_{i=1}^{n} a_{i}, B=\sum_{i=1}^{n} b_{i}, the problem is equivalent to proving: (ABi=1naibi)2(A2\left(A B-\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \geqslant\left(A^{2}-\right. i=1nai2)(B2i=1nbi2)\left.\sum_{i=1}^{n} a_{i}^{2}\right)\left(B^{2}-\sum_{i=1}^{n} b_{i}^{2}\right)

Construct a quadratic function: f(x)=(A2i=1nai2)x2+2(ABi=1naibi)x+(B2i=1nbi2)f(x)=\left(A^{2}-\sum_{i=1}^{n} a_{i}^{2}\right) x^{2}+2\left(A B-\sum_{i=1}^{n} a_{i} b_{i}\right) x+\left(B^{2}-\sum_{i=1}^{n} b_{i}^{2}\right).
We need to prove that its discriminant Δ0\Delta \geqslant 0, now we just need to verify that there exists a real number x0x_{0} such that f(x0)0f\left(x_{0}\right) \leqslant 0, take
x0=BA,f(BA)0B2i=1nai22ABi=1naibi+A2i=1nbi20x_{0}=-\frac{B}{A}, f\left(-\frac{B}{A}\right) \leqslant 0 \Leftrightarrow B^{2} \sum_{i=1}^{n} a_{i}^{2}-2 A B \sum_{i=1}^{n} a_{i} b_{i}+A^{2} \sum_{i=1}^{n} b_{i}^{2} \geqslant 0

To prove (1), we just need to prove that its discriminant Δ10\Delta_{1} \leqslant 0, which is obtained by the Cauchy-Schwarz inequality (i=1naibi)2\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \leqslant i=1nai2i=1nbi2\sum_{i=1}^{n} a_{i}^{2} \sum_{i=1}^{n} b_{i}^{2}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.