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Difficulty 6.6 National olympiad Prove it
124. For any three distinct positive real numbers a , b , c a, b, c a , b , c ( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 > 8 a b c \frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}>8 a b c ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 ( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 > 8 ab c
Solution 124. By factorization we get( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 = 3 ( a − b ) ( b − c ) ( c − a ) ( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 = 3 ( a 2 − b 2 ) ( b 2 − c 2 ) ( c 2 − a 2 ) \begin{aligned}
(a-b)^{3}+(b-c)^{3}+(c-a)^{3} & =3(a-b)(b-c)(c-a) \\
\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3} & =3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)
\end{aligned} ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 ( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 = 3 ( a − b ) ( b − c ) ( c − a ) = 3 ( a 2 − b 2 ) ( b 2 − c 2 ) ( c 2 − a 2 )
Therefore,( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 = ( a + b ) ( b + c ) ( c + a ) > 8 a b c \frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}=(a+b)(b+c)(c+a)>8 a b c ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 ( a 2 − b 2 ) 3 + ( b 2 − c 2 ) 3 + ( c 2 − a 2 ) 3 = ( a + b ) ( b + c ) ( c + a ) > 8 ab c
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