Library / Ukrajina / 1 of 23
Ukrajina, 2008 Difficulty 4.5 AIME Prove it Ukraine
Prove inequality a 2 b c ( a 2 + b 2 ) + b 2 c a ( b 2 + c 2 ) + c 2 a b ( c 2 + a 2 ) ≥ 9 2 \frac{a^2}{bc(a^2+b^2)} + \frac{b^2}{ca(b^2+c^2)} + \frac{c^2}{ab(c^2+a^2)} \ge \frac{9}{2} b c ( a 2 + b 2 ) a 2 + c a ( b 2 + c 2 ) b 2 + ab ( c 2 + a 2 ) c 2 ≥ 2 9 a , b , c a, b, c a , b , c a b + b c + c a = 1 ab + bc + ca = 1 ab + b c + c a = 1
Solution Let's consider the secondary inequality a 3 a 2 + b 2 ≥ a − b 2 \frac{a^3}{a^2+b^2} \ge a - \frac{b}{2} a 2 + b 2 a 3 ≥ a − 2 b 2 a 3 ≥ ( a 2 + b 2 ) ( 2 a − b ) = 2 a 3 + 2 a b 2 − a 2 b − b 3 ⇔ b ( a − b ) 2 ≥ 0 2a^3 \ge (a^2+b^2)(2a-b) = 2a^3 + 2ab^2 - a^2b - b^3 \Leftrightarrow b(a-b)^2 \ge 0 2 a 3 ≥ ( a 2 + b 2 ) ( 2 a − b ) = 2 a 3 + 2 a b 2 − a 2 b − b 3 ⇔ b ( a − b ) 2 ≥ 0
a 2 b c ( a 2 + b 2 ) + b 2 c a ( b 2 + c 2 ) + c 2 a b ( c 2 + a 2 ) = 1 a b c ( a 3 a 2 + b 2 + b 3 b 2 + c 2 + c 3 c 2 + a 2 ) ≥ 1 a b c ( a − b 2 + b − c 2 + c − a 2 ) = a + b + c 2 a b c = ( a + b + c ) ( a b + b c + c a ) 2 a b c ≥ 3 a b c ⋅ a 2 b 2 c 2 3 2 a b c ≥ 9 2 . \frac{a^2}{bc(a^2+b^2)} + \frac{b^2}{ca(b^2+c^2)} + \frac{c^2}{ab(c^2+a^2)} = \frac{1}{abc} \left( \frac{a^3}{a^2+b^2} + \frac{b^3}{b^2+c^2} + \frac{c^3}{c^2+a^2} \right) \ge \frac{1}{abc} \left( a - \frac{b}{2} + b - \frac{c}{2} + c - \frac{a}{2} \right) = \frac{a+b+c}{2abc} = \frac{(a+b+c)(ab+bc+ca)}{2abc} \ge \frac{3\sqrt{abc} \cdot \sqrt[3]{a^2b^2c^2}}{2abc} \ge \frac{9}{2}. b c ( a 2 + b 2 ) a 2 + c a ( b 2 + c 2 ) b 2 + ab ( c 2 + a 2 ) c 2 = ab c 1 ( a 2 + b 2 a 3 + b 2 + c 2 b 3 + c 2 + a 2 c 3 ) ≥ ab c 1 ( a − 2 b + b − 2 c + c − 2 a ) = 2 ab c a + b + c = 2 ab c ( a + b + c ) ( ab + b c + c a ) ≥ 2 ab c 3 ab c ⋅ 3 a 2 b 2 c 2 ≥ 2 9 .
First Browse the library Next →
Want a route through all this instead of an archive?
The track
puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.
Source: MathNet ,
licensed CC-BY-4.0 .
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.