Track / Stage 6 / 68 of 400 #1068 of 2000
Problem 1068 National olympiad, first round Difficulty 6.1 Prove it
16. Real numbers x , y , z x, y, z x , y , z x y z = 1 x y z=1 x y z = 1 x 2 ( x − 1 ) 2 + y 2 ( y − 1 ) 2 + z 2 ( z − 1 ) 2 ⩾ 1 \frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geqslant 1 ( x − 1 ) 2 x 2 + ( y − 1 ) 2 y 2 + ( z − 1 ) 2 z 2 ⩾ 1
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Official solution 16. x 2 ( x − 1 ) 2 + y 2 ( y − 1 ) 2 + z 2 ( z − 1 ) 2 − 1 ≡ a 6 ( a 3 − a b c ) 2 + b 6 ( b 3 − a b c ) 2 + c 6 ( c 3 − a b c ) 2 − 1 = ( b c + c a + a b ) 2 ( b 2 c 2 + c 2 a 2 + a 2 b 2 − a 2 b c − b 2 c a − c 2 a b ) 2 ( a 2 − b c ) 2 ( b 2 − c a ) 2 ( c 2 − a b ) 2 ⩾ 0 \begin{aligned} & \frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}}-1 \\ \equiv & \frac{a^{6}}{\left(a^{3}-a b c\right)^{2}}+\frac{b^{6}}{\left(b^{3}-a b c\right)^{2}}+\frac{c^{6}}{\left(c^{3}-a b c\right)^{2}}-1 \\ = & \frac{(b c+c a+a b)^{2}\left(b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}-a^{2} b c-b^{2} c a-c^{2} a b\right)^{2}}{\left(a^{2}-b c\right)^{2}\left(b^{2}-c a\right)^{2}\left(c^{2}-a b\right)^{2}} \geqslant 0\end{aligned} ≡ = ( x − 1 ) 2 x 2 + ( y − 1 ) 2 y 2 + ( z − 1 ) 2 z 2 − 1 ( a 3 − ab c ) 2 a 6 + ( b 3 − ab c ) 2 b 6 + ( c 3 − ab c ) 2 c 6 − 1 ( a 2 − b c ) 2 ( b 2 − c a ) 2 ( c 2 − ab ) 2 ( b c + c a + ab ) 2 ( b 2 c 2 + c 2 a 2 + a 2 b 2 − a 2 b c − b 2 c a − c 2 ab ) 2 ⩾ 0
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Source: NuminaMath-1.5 ,
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