An algebra problem by Anish Roy

Algebra Level 4

Suppose ( x 1 , x 2 , ⋯ , x n , ⋯ ) (x_{1},x_{2}, \cdots ,x_{n}, \cdots ) is a sequence of positive real numbers such that x 1 ≥ x 2 ≥ x 3 ≥ ⋯ ≥ x n ⋯ x_{1} \geq x_{2} \geq x_{3} \geq \cdots \geq x_{n} \cdots , and for all n n .

x 1 1 + x 4 2 + x 9 3 + ⋯ + x n 2 n ≤ 1 \frac {x_{1}}{1} + \frac {x_{4}}{2} + \frac {x_{9}}{3} + \cdots + \frac {x_{n^2}}{n} \leq 1

Then find the maximum value of

x 1 1 + x 2 2 + x 3 3 + ⋯ + x k k \frac {x_{1}}{1} + \frac {x_{2}}{2} + \frac {x_{3}}{3} + \cdots + \frac {x_{k}}{k}


The answer is 3.

Anish Roy by Anish Roy , 25, India

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1 solution

Anish Roy
Aug 10, 2017

Let k k be a natural number and let n n be the unique integer such that ( n − 1 ) 2 ≤ k < n 2 (n-1)^2 \leq k < n^2 Then we see that ∑ r = 1 k x r r ≤ ( x 1 1 + x 2 2 + x 3 3 ) + ( x 4 4 + x 5 5 + ⋯ + x 8 8 ) + ⋯ + ( x ( n − 1 ) 2 ( n − 1 ) 2 + ⋯ + x k k + ⋯ + x n 2 − 1 n 2 − 1 ) \displaystyle \sum_{r=1}^k \frac {x_{r}}{r} \leq (\frac {x_{1}}{1}+\frac {x_{2}}{2}+\frac {x_{3}}{3})+(\frac {x_{4}}{4}+\frac {x_{5}}{5}+ \cdots +\frac {x_{8}}{8})+ \cdots + (\frac {x_{(n-1)^2}}{(n-1)^2}+ \cdots +\frac {x_{k}}{k}+ \cdots +\frac {x_{n^2-1}}{n^2-1}) ≤ ( x 1 1 + x 1 1 + x 1 1 ) + ( x 4 4 + x 4 4 + ⋯ + x 4 4 ) + ⋯ + ( x ( n − 1 ) 2 ( n − 1 ) 2 + ⋯ + x ( n − 1 ) 2 ( n − 1 ) 2 ) \leq (\frac {x_{1}}{1}+\frac {x_{1}}{1}+\frac {x_{1}}{1})+(\frac {x_{4}}{4}+\frac {x_{4}}{4}+ \cdots +\frac {x_{4}}{4})+ \cdots + (\frac {x_{(n-1)^2}}{(n-1)^2}+ \cdots +\frac {x_{(n-1)^2}}{(n-1)^2}) = 3 x 1 1 + 5 x 2 4 + ⋯ + ( 2 n − 1 ) x ( n − 1 ) 2 ( n − 1 ) 2 = \frac {3x_{1}}{1}+\frac {5x_{2}}{4}+\cdots+\frac {(2n-1)x_{(n-1)^2}}{(n-1)^2} = ∑ r = 1 n − 1 ( 2 r + 1 ) x r 2 r 2 = \displaystyle \sum_{r=1}^{n-1} \frac {(2r+1)x_{r^2}}{r^2} ≤ ∑ r = 1 n − 1 ( 3 r ) x r 2 r 2 \leq \displaystyle \sum_{r=1}^{n-1} \frac {(3r)x_{r^2}}{r^2} = 3 ∑ r = 1 n − 1 x r 2 r = 3\displaystyle \sum_{r=1}^{n-1} \frac {x_{r^2}}{r} ≤ 3 \leq 3 where the last inequality follows from the given hypothesis.

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