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Difficulty 5.5 AIME, harder Prove it
Example 1. If a , b , c a, b, c a , b , c 1 ( a − b ) 2 + 1 ( b − c ) 2 + 1 ( c − a ) 2 \sqrt{\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}} ( a − b ) 2 1 + ( b − c ) 2 1 + ( c − a ) 2 1
Solution Prove ∵ ( a − b ) + ( b − c ) + ( c − a ) = 0 \because(a-b)+(b-c)+(c-a)=0 ∵ ( a − b ) + ( b − c ) + ( c − a ) = 0 ∴ 1 ( a − b ) 2 + 1 ( b − c ) 2 + 1 ( c − a ) 2 = ( 1 a − b + 1 b − c + 1 c − a ) 2 . ∴ 1 ( a − b ) 2 + 1 ( b − c ) 2 + 1 ( c − a ) 2 = ∣ 1 a − b + 1 b − c + 1 i − a ∣ .
\begin{aligned}
\therefore & \frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}} \\
& =\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^{2} . \\
\therefore \quad & \sqrt{\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}} \\
& =\left|\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{i-a}\right| .
\end{aligned}
∴ ∴ ( a − b ) 2 1 + ( b − c ) 2 1 + ( c − a ) 2 1 = ( a − b 1 + b − c 1 + c − a 1 ) 2 . ( a − b ) 2 1 + ( b − c ) 2 1 + ( c − a ) 2 1 = a − b 1 + b − c 1 + i − a 1 .
The conclusion is obvious.
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