Track / Stage 7 / 144 of 300 #1544 of 1964
Problem 1544 National Olympiad second round; IMO P1/P4 Difficulty 7.3 Prove it
Let { b n } \left\{b_{n}\right\} { b n } n ⩾ 1 n \geqslant 1 n ⩾ 1 b n + 1 2 ⩾ b 1 2 1 3 + b 2 2 2 3 + ⋯ + b n 2 n 3 b_{n+1}^{2} \geqslant \frac{b_{1}^{2}}{1^{3}}+\frac{b_{2}^{2}}{2^{3}}+\cdots+\frac{b_{n}^{2}}{n^{3}} b n + 1 2 ⩾ 1 3 b 1 2 + 2 3 b 2 2 + ⋯ + n 3 b n 2 k k k ∑ n = 1 k b n + 1 b 1 + b 2 + ⋯ + b n > 1993 1000 \sum_{n=1}^{k} \frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}}>\frac{1993}{1000} ∑ n = 1 k b 1 + b 2 + ⋯ + b n b n + 1 > 1000 1993 x 1 , x 2 , ⋯ , x 2001 x_{1}, x_{2}, \cdots, x_{2001} x 1 , x 2 , ⋯ , x 2001 x i 2 ⩾ x 1 2 1 3 + x 2 2 2 3 + ⋯ + x i − 1 2 ( i − 1 ) 3 , 2 ⩽ i ⩽ 2001 x_{i}^{2} \geqslant \frac{x_{1}^{2}}{1^{3}}+\frac{x_{2}^{2}}{2^{3}}+\cdots+\frac{x_{i-1}^{2}}{(i-1)^{3}}, 2 \leqslant i \leqslant 2001 x i 2 ⩾ 1 3 x 1 2 + 2 3 x 2 2 + ⋯ + ( i − 1 ) 3 x i − 1 2 , 2 ⩽ i ⩽ 2001 ∑ i = 2 2001 x i x 1 + x 2 + ⋯ + x i − 1 > 1999 \sum_{i=2}^{2001} \frac{x_{i}}{x_{1}+x_{2}+\cdots+x_{i-1}}>1999 ∑ i = 2 2001 x 1 + x 2 + ⋯ + x i − 1 x i > 1999
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Official solution 10. (1) By the Cauchy inequality, we have ( 1 3 + 2 3 + ⋯ + n 3 ) b n + 1 2 ⩾ ( 1 3 + 2 3 + ⋯ + n 3 ) \left(1^{3}+2^{3}+\cdots+n^{3}\right) b_{n+1}^{2} \geqslant\left(1^{3}+2^{3}+\cdots+n^{3}\right) ( 1 3 + 2 3 + ⋯ + n 3 ) b n + 1 2 ⩾ ( 1 3 + 2 3 + ⋯ + n 3 ) ( b 1 2 1 3 + b 2 2 2 3 + ⋯ + b n 2 n 3 ) = ( b 1 + b 2 + ⋯ + b n ) 2 \left(\frac{b_{1}^{2}}{1^{3}}+\frac{b_{2}^{2}}{2^{3}}+\cdots+\frac{b_{n}^{2}}{n^{3}}\right)=\left(b_{1}+b_{2}+\cdots+b_{n}\right)^{2} ( 1 3 b 1 2 + 2 3 b 2 2 + ⋯ + n 3 b n 2 ) = ( b 1 + b 2 + ⋯ + b n ) 2
Since 1 3 + 2 3 + ⋯ + n 3 = ( n ( n + 1 ) 2 ) 2 1^{3}+2^{3}+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2} 1 3 + 2 3 + ⋯ + n 3 = ( 2 n ( n + 1 ) ) 2 b n + 1 b 1 + b 2 + ⋯ + b n ⩾ 2 n ( n + 1 ) = 2 n − 2 n + 1 \frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}} \geqslant \frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1} b 1 + b 2 + ⋯ + b n b n + 1 ⩾ n ( n + 1 ) 2 = n 2 − n + 1 2
Therefore,∑ n = 1 k b n + 1 b 1 + b 2 + ⋯ + b n ⩾ ∑ n = 1 k ( 2 n − 2 n + 1 ) = 2 ( 1 − 1 k + 1 ) \sum_{n=1}^{k} \frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}} \geqslant \sum_{n=1}^{k}\left(\frac{2}{n}-\frac{2}{n+1}\right)=2\left(1-\frac{1}{k+1}\right) n = 1 ∑ k b 1 + b 2 + ⋯ + b n b n + 1 ⩾ n = 1 ∑ k ( n 2 − n + 1 2 ) = 2 ( 1 − k + 1 1 )
Thus, by taking k = 999 k=999 k = 999 2 ( 1 − 1 1000 ) = 1998 1000 > 1993 1000 2\left(1-\frac{1}{1000}\right)=\frac{1998}{1000}>\frac{1993}{1000} 2 ( 1 − 1000 1 ) = 1000 1998 > 1000 1993 k = 2001 k=2001 k = 2001
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