Wow. Such summation. Much variables. Very degrees.

Algebra Level pending

If the minimum value of c c that satisfy the inequality for all x 1 , x 2 , … , x n ≥ 0 x_{1},x_{2},\dots,x_{n} \geq 0

∑ 1 ≤ i < j ≤ n x i x j ( x i 2 + x j 2 ) ≤ c ( ∑ 1 ≤ i ≤ n x i ) 4 \displaystyle \large \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leq c \left(\sum\limits_{1 \leq i \leq n} x_{i}\right)^{4}

can be expressed as a b \displaystyle \frac{a}{b} for coprime integers a , b \large a,b . What is the value of a 3 + b 3 a^{3}+b^{3} ?

If c < 0 \displaystyle c <0 , then a < 0 \displaystyle a < 0 and b > 0 \displaystyle b > 0 .

  • This problem is not original.


The answer is 513.

Samuraiwarm Tsunayoshi by Samuraiwarm Tsunayoshi , 22, Thailand

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1 solution

Add all the remaining terms from 1 1 to n n in the bracket from LHS

∑ 1 ≤ i < j ≤ n x i x j ( x i 2 + x j 2 ) \displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)

≤ ∑ 1 ≤ i < j ≤ n x i x j ( x 1 2 + x 2 2 + ⋯ + x n 2 ) \leq \displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}\left(x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2}\right)

= ( x 1 2 + x 2 2 + . . . + x n 2 ) ∑ 1 ≤ i < j ≤ n x i x j = (x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2})\displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}

Consider ( x 1 + x 2 + . . . + x n ) 2 = ( x 1 2 + x 2 2 + . . . + x n 2 ) + 2 ∑ 1 ≤ i < j ≤ n x i x j \left(x_{1}+x_{2}+...+x_{n}\right)^{2} = (x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}) + 2 \displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}

≥ 2 ( x 1 2 + x 2 2 + . . . + x n 2 ) × 2 ∑ 1 ≤ i < j ≤ n x i x j \geq 2\sqrt{(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}) \times 2 \displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}} by AM-GM.

Square both sides we get

( x 1 + x 2 + . . . + x n ) 4 ≥ 8 ( x 1 2 + x 2 2 + . . . + x n 2 ) ∑ 1 ≤ i < j ≤ n x i x j ≥ 8 ∑ 1 ≤ i < j ≤ n x i x j ( x i 2 + x j 2 ) \displaystyle \left(x_{1}+x_{2}+...+x_{n}\right)^{4} \geq 8(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2})\displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j} \geq 8\displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)

We get

∑ 1 ≤ i < j ≤ n x i x j ( x i 2 + x j 2 ) ≤ 1 8 ( ∑ 1 ≤ i ≤ n x i ) 4 \displaystyle \sum\limits_{1 \leq i < j \leq n} x_{i}x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leq \displaystyle \frac{1}{8} \left(\sum\limits_{1 \leq i \leq n} x_{i}\right)^{4}

Hence, c = 1 8 \displaystyle c = \boxed{\displaystyle \frac{1}{8}} ~~~

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