Track / Stage 6 / 167 of 400 #1167 of 2000
Problem 1167 National olympiad, first round Difficulty 6.3 Prove it
Example 5 Let a , b , c a, b, c a , b , c ( a − b b − c ) 2 + ( b − c c − a ) 2 + ( c − a a − b ) 2 ⩾ 5 \left(\frac{a-b}{b-c}\right)^{2}+\left(\frac{b-c}{c-a}\right)^{2}+\left(\frac{c-a}{a-b}\right)^{2} \geqslant 5 ( b − c a − b ) 2 + ( c − a b − c ) 2 + ( a − b c − a ) 2 ⩾ 5
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.
I solved it I didn't Skip
Official solution Prove ( a − b b − c ) 2 + ( b − c c − a ) 2 + ( c − a a − b ) 2 = 5 + ( 1 + a − b b − c + b − c c − a + c − a a − b ) 2 ⩾ 5. \quad \begin{aligned} & \left(\frac{a-b}{b-c}\right)^{2}+\left(\frac{b-c}{c-a}\right)^{2}+\left(\frac{c-a}{a-b}\right)^{2} \\ = & 5+\left(1+\frac{a-b}{b-c}+\frac{b-c}{c-a}+\frac{c-a}{a-b}\right)^{2} \geqslant 5 .\end{aligned} = ( b − c a − b ) 2 + ( c − a b − c ) 2 + ( a − b c − a ) 2 5 + ( 1 + b − c a − b + c − a b − c + a − b c − a ) 2 ⩾ 5.
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