Maths Olympiad Prep

Library / /444 of 520

Algebra Difficulty 7.4 National olympiad, round 2 Prove it

Example 11\mathbf{1 1} (2006 Korean Olympiad Problem) For three distinct real numbers a1,a2,a3a_{1}, a_{2}, a_{3}, define three real numbers b1,b2,b3b_{1}, b_{2}, b_{3} as follows: bj=(1+ajaiaj−ai)(1+ajakaj−ak)b_{j}=\left(1+\frac{a_{j} a_{i}}{a_{j}-a_{i}}\right)\left(1+\frac{a_{j} a_{k}}{a_{j}-a_{k}}\right), where {i,j,k}={1,2,3}\{i, j, k\}=\{1,2,3\}. Prove that 1+∣a1b1+a2b2+a3b3∣⩽(1+∣a1∣)(1+∣a2∣)(1+∣a3∣)1+\left|a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right| \leqslant\left(1+\left|a_{1}\right|\right)\left(1+\left|a_{2}\right|\right)\left(1+\left|a_{3}\right|\right), and determine the conditions under which equality holds.

Solution

Proof: Let A=a1a2a1−a2,B=a1a3a1−a3,C=a2a3a2−a3 A = \frac{a_1 a_2}{a_1 - a_2}, B = \frac{a_1 a_3}{a_1 - a_3}, C = \frac{a_2 a_3}{a_2 - a_3} , then
a1b1+a2b2+a3b3=a1(1+A)(1+B)+a2(1−A)(1+C)+a3(1−B)(1−C)=a1+a2+a3+(a1−a2)A+(a1−a3)B+(a2−a3)C+a1AB−a2AC+a3BC \begin{aligned} & a_1 b_1 + a_2 b_2 + a_3 b_3 = a_1 (1 + A)(1 + B) + a_2 (1 - A)(1 + C) + a_3 (1 - B)(1 - C) \\ = & a_1 + a_2 + a_3 + (a_1 - a_2) A + (a_1 - a_3) B + (a_2 - a_3) C + a_1 A B - a_2 A C + a_3 B C \end{aligned}

By calculation, we get (a1−a2)A+(a1−a3)B+(a2−a3)C=a1a2+a1a3+a2a3(a_1 - a_2) A + (a_1 - a_3) B + (a_2 - a_3) C = a_1 a_2 + a_1 a_3 + a_2 a_3,
a1AB−a2AC+a3BC=a1a2a3a12(a2−a3)+a22(a3−a1)+a32(a1−a2)(a1−a2)(a2−a3)(a1−a3)=a1a2a3. a_1 A B - a_2 A C + a_3 B C = a_1 a_2 a_3 \frac{a_1^2 (a_2 - a_3) + a_2^2 (a_3 - a_1) + a_3^2 (a_1 - a_2)}{(a_1 - a_2)(a_2 - a_3)(a_1 - a_3)} = a_1 a_2 a_3.

Therefore, 1+∣a1b1+a2b2+a3b3∣=1+∣a1+a2+a3+a1a2+a1a3+a2a3+a1a2a3∣1 + |a_1 b_1 + a_2 b_2 + a_3 b_3| = 1 + |a_1 + a_2 + a_3 + a_1 a_2 + a_1 a_3 + a_2 a_3 + a_1 a_2 a_3|
≤1+∣a1∣+∣a2∣+∣a3∣+∣a1a2∣+∣a1a3∣+∣a2a3∣+∣a1a2a3∣=(1+∣a1∣)(1+∣a2∣)(1+∣a3∣). \begin{array}{l} \leq 1 + |a_1| + |a_2| + |a_3| + |a_1 a_2| + |a_1 a_3| + |a_2 a_3| + |a_1 a_2 a_3| \\ = (1 + |a_1|)(1 + |a_2|)(1 + |a_3|). \end{array}

The equality holds if and only if the seven real numbers a1,a2,a3,a1b1,a2b2,a3b3,a1a2a3a_1, a_2, a_3, a_1 b_1, a_2 b_2, a_3 b_3, a_1 a_2 a_3 are all non-negative or all non-positive. Note that at most one of a1,a2,a3a_1, a_2, a_3 is 0, thus the equality holds if and only if a1,a2,a3a_1, a_2, a_3 are all non-negative.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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